Axioms, definitions and theorems
Every proof answers “why?” with an earlier fact, in words that must mean one exact thing. Where do the first facts and the exact meanings come from?
Where you are 14 in Proof techniques
Why this exists
A proof gives a reason for a statement, and the reason is an earlier statement. Ask “why?” of that one, and you are handed an even earlier one. The questions cannot run backwards for ever, and they cannot go round in a circle, because a circle proves nothing. Somewhere the answers have to stop.
The statements where they stop are called axioms. Choosing them openly is what lets mathematics say exactly what it is talking about. Euclid built the geometry of the plane on a short list of starting statements, and every theorem in it can be traced back to that list. In the nineteenth century people changed one item on the list and found a second, different geometry that works just as well as the first. Around 1900 a harsher lesson arrived in the theory of sets: a rule that sounded perfectly obvious, that any description picks out a set, turned out to contradict itself. Both stories say the same thing. Whatever you start from has to be stated and examined, because you cannot see what a proof depends on unless you can see where it starts.
The words in a proof need the same care. Two people argue about whether 0 is even. One says it cannot be, because even numbers are the ones you can share out in pairs, and there is nothing to share. The other says it must be, because it sits between -1 and 1, and both of those are odd. They can go on all evening. Nothing either of them says can settle it, because each is using the word “even” to mean a slightly different thing. The cure is to agree, in words that leave no room, on exactly what the word means. After that the argument is over, because it is no longer an argument about words. You hold the agreed meaning up against the number and look. That is what a definition is for. It turns a word that has a feel into a rule that has a verdict. “Rectangle”, “prime”, “continuous” and “derivative” are each, before anything is proved about them, one exact sentence, and every proof about them begins from that sentence. A proof has no way of knowing what a word “really” means. It knows only what the word was defined to mean, and that is all it needs.
What comes out of axioms and definitions are proved statements, and books label them: Theorem 2.3, Lemma 2.2, Corollary 2.4. All three kinds have proofs, and all three are true for the same kind of reason, so the labels are not telling the reader how sure to be. They say what a statement is for. A proof of anything large is hard to read as one long wall of reasoning, so it is cut into pieces: smaller facts proved first, then the main result, then the easy things that fall out of it. Each piece has a different job, and the three words name the job. Without them a long text is a pile of true statements in a row. With them, the reader can see what leads to what.
Definition
A mathematical system is built from three kinds of statement. Axioms are accepted without proof. Definitions fix what new words mean. Theorems are proved from the axioms and the definitions, and among proved statements the words lemma and corollary name particular jobs. This section takes the three in turn.
What an axiom is
An axiom is a statement that a mathematical system accepts without proof and uses as a starting point. Three things make a statement an axiom:
- it is stated before any proving begins,
- any proof in the system is allowed to use it as a reason,
- the system does not prove it. The system chooses it.
The older word postulate names the same thing, and many geometry books still use it. An axiom usually talks about objects that the system leaves undefined, such as “point” and “line” (see undefined terms). The axioms are then the only thing the system knows about those objects, and everything else has to come from them.
A statement that is proved from the axioms is a theorem. The difference between the two is only where they sit: a theorem has reasons beneath it, and an axiom is where the reasons end.
Here is a system small enough to hold in your head. It has three axioms, about two kinds of thing, points and lines.
Axiom 1. There are exactly three points.
Axiom 2. Any two different points lie on exactly one line.
Axiom 3. Every line has exactly two points on it.
Nothing else is known. A line here is not straight or long, and not even drawn. It is whatever the axioms say it is. Figure 1 is one picture of a world that obeys all three.
What a definition does
A definition introduces a word by stating, in words already understood, the exact condition something must meet to be called by it.
Even. An integer n is even if n = 2k for some integer k.
The sentence has three parts: the word being defined (even), the kind of thing it applies to (an integer n), and the condition (there is an integer k with n = 2k). Geometry’s words work the same way.
Rectangle. A rectangle is a quadrilateral with four right angles.
Here the word is rectangle, the kind of thing is quadrilateral, and the condition is four right angles.
Three features are worth knowing before you use one.
It uses only words already understood. In the definition of even, the words “integer” and “times” were settled earlier. Every word inside a definition has to have been defined before, or else be one of the few terms a subject accepts without definition, such as the “point” and “line” of the axiom system above.
“If” means “if and only if”. The definition is written “is even if n = 2k”, but it means more than that: an integer is even when n = 2k for some integer k, and only then. Definitions are usually written with a bare “if” and always read as a two-way statement. Why that reading is forced, and what it has to do with Biconditional statements, is taken up in converses and biconditional definitions.
It is not true or false. A definition does not report a fact that might turn out wrong. It is a decision about how a word will be used. There is nothing to prove about it. You can accept it, or you can choose a different one, and nobody has made a mistake either way.
That is how a definition differs from an axiom, though neither is proved. An axiom makes a claim about its objects, so a given world can keep it or break it, as the one-line world of Example 6 breaks Axiom 3. A definition makes no claim at all. It only gives a name to a condition, so no world can break it.
Theorem, lemma and corollary
There are three words for proved statements, and each names a job.
A theorem is a proved statement that a text presents as one of its main results: worth stating, remembering and citing for its own sake.
A lemma is a proved statement that is there chiefly to help prove something else. It is a stepping stone.
A corollary is a proved statement that follows quickly from a result just proved, often a special case of it or a different way of saying it. Its proof is short, because the work has already been done: cite the result and add a line or two.
Many books also use proposition for a proved statement that carries less weight than a theorem. Where the line falls is the author’s choice. Older usage differs again: the usual English editions of Euclid’s Elements call every numbered result a proposition, great or small, as in Book I, Proposition 15.
The first part of this section used the word theorem for any statement proved from the axioms. That is the wide meaning, and it still holds: a lemma and a corollary are theorems in the wide sense, because each is proved. The word in the list above is the narrow meaning, a label for the main results. When this lesson writes “theorem” beside “lemma” and “corollary”, it means the narrow one.
Here is one of each, all about integers. They are proved in Examples 16 to 18.
Lemma. Every square of an integer leaves remainder 0 or 1 when divided by 4.
Theorem. A sum of two squares of integers never leaves remainder 3 when divided by 4.
Corollary. The equation x^2 + y^2 = 2027 has no solution in integers.
The lemma is there because the theorem needs it. The theorem is the result worth having. The corollary is one instance of the theorem: 2027 leaves remainder 3 when divided by 4, so it is one of the numbers the theorem rules out.
Intuition
Axioms as the rules of a game
Think of the axioms as the rules of a game. A rule of chess is not true or false; it is what makes the game that game. Once the rules are fixed, a statement counts as true in the system exactly when it follows from them, and nothing else counts. Someone else may play a game with a different rule, and that is not a mistake by either player. It is a different game.
Figure 2 shows what “follows from” looks like. Each theorem stands on earlier statements. Follow any arrow down and you reach an axiom, and no arrow leads into an axiom, because an axiom is not supported by anything in the system.
There is a second picture worth keeping beside this one. A set of axioms describes a kind of world: any collection of things, with whatever the axioms mention, in which every axiom comes out true. Figure 1 is a world for the three axioms above. One set of axioms can have many worlds: Axioms 2 and 3 on their own hold in the three-point world and also in a world of four points (Example 2). The axioms of Euclid’s plane hold in the plane built out of pairs of real numbers, which a flat sheet of paper pictures well, but a different set of axioms is needed for the surface of a ball. A drawing of a world is only a picture of it. What the world contains is settled by the axioms, not by how the drawing looks.
A definition as a membership test
A definition is a membership test, written out in full. The word names everyone who passes the test and nobody else.
Here is the test for even, as a picture. Stand at 0 on a number line and hop by 2. After k hops to the right you land on 2k, and after k hops to the left you land on -2k. Saying “n = 2k for some integer k” says that n is one of the landing points.
The landing points are \dots, -4, -2, 0, 2, 4, 6, \dots. The number 7 is not one of them: three hops reach 6 and four reach 8, so 7 is jumped over. And 0 is one, because it is where the hopping starts: zero hops, k = 0, and 0 = 2 \cdot 0. The old argument about zero is settled without any argument, by running the test.
The test also works in both directions. If you are told a number is a landing point, you may write it as 2k. If you can write it as 2k, you may say it is a landing point. That two-way freedom is what makes a definition so useful inside a proof.
Theorems, lemmas and corollaries as jobs
Figure 4 is built like Figure 2, but each proved statement’s box now names its job. It shows a small text arranged the way the three words suggest. Read it from the bottom.
Two lemmas are proved first, each straight from the axioms and definitions. The theorem is then proved from the two lemmas: its proof is short because the lemmas did the heavy lifting. Two corollaries are proved last, each in a line or two from the theorem. An arrow always points from a reason to what it supports, and every chain of arrows ends at the axioms.
Lemma and corollary describe a statement’s place beside another statement. We say “Lemma 1 is a lemma for Theorem 3”, and “Corollary 4 is a corollary of Theorem 3”. So the same statement can be both at once. In Example 14 a result about quadrilaterals is a corollary of the triangle angle sum theorem, and then in Example 15 it is used as a lemma for a result about pentagons.
From the writer’s side, here is when a step is worth pulling out as a lemma.
- It is long, and giving it a name and its own proof keeps the main argument short enough to follow.
- It is needed more than once. Proving it a single time beats repeating the same work in every place it is used.
- It is interesting on its own. Then it may deserve to be remembered, and the name follows the importance.
A corollary is the reverse move. After a big result, the immediate consequences are written down once, in plain view, so the reader sees what the result buys without having to rediscover it.
The labels are also instructions for reading. A theorem is what to take away. A lemma can be skipped on a first pass, and trusted until you want to know why it is true. A corollary is worth a moment to check that it really is a quick step from its parent.
Results
What an axiom is
A theorem is a statement reached from the axioms by valid steps. In a system, “true” means exactly this, and the steps are the reasons of deductive reasoning.
For a fixed set of axioms that does not contradict itself, every statement gets exactly one of three verdicts.
- provable: a proof from the axioms exists,
- refutable: the opposite statement is provable,
- undecided: neither the statement nor its opposite can be proved from these axioms.
The verdict belongs to the pair (statement, axioms). Change the axioms and the verdict can change.
To show that a statement cannot be proved, build a world. If there is a world in which every axiom is true and the statement is false, then no proof of the statement from those axioms exists.
A usable set of axioms cannot contradict itself. If a statement and its opposite are both provable, there is no world at all in which the axioms hold.
What a definition does: unfold and fold
Using a definition. Suppose a word W is defined by a condition C. Then two moves are always allowed, and neither needs any reason except the definition itself.
- Unfold. If something is a W, then it meets C.
- Fold. If something meets C, then it is a W.
For even, unfolding turns “n is even” into “n = 2k for some integer k”. Folding goes back the other way: from “n = 2m for some integer m” it reaches “n is even”.
Two habits keep the moves honest.
An unfolded “some” gives a name, not a number you chose. When the definition says “for some integer k”, unfolding hands you an integer, which you call k, and all you know about it is the equation. Each separate unfolding gets its own letter. If m and n are both even, write m = 2a and n = 2b. Writing 2k for both would quietly say that m = n.
Folding means producing exactly what the definition asks for. To show a number is even you must show it equals 2 times an integer. Write the number as 2 times something, then check that the something really is an integer.
A proof that uses a defined word then has a simple shape: unfold what you are given, do ordinary algebra or reasoning, and fold to reach what you must show. This is the frame inside every Direct proof.
Theorem, lemma and corollary
A theorem, a lemma and a corollary are all proved, and none is less certain than another. Each follows from the axioms and from statements proved earlier by valid steps. The word says what the statement is for, not how certain it is.
Any of the three may be cited as a reason. Once a lemma or a corollary is proved, a later proof can use it exactly as it would use a theorem.
The words add no logical power. A proof that cites a lemma can be rewritten so that it cites no lemma, by writing the lemma’s own proof in its place. The same goes for corollaries. The names organize a proof. They do not make it stronger.
Nothing may be cited before it is proved, and there are no circles. If a lemma is used to prove a theorem, then the proof of the lemma must not use that theorem, directly or through other statements.
A flaw travels upward. If the proof of a statement has a gap, every statement that stands on it is left without a proof until the gap is mended, and nothing that stands beneath it is affected. A corollary is only as secure as its parent, and a theorem is only as secure as its lemmas.
Why it is true
What an axiom is
Why the reasons must stop. A proof is a finite list of lines, and each line is justified by lines above it or by statements accepted without proof. Look at the first line. Nothing is above it, so the only thing that can justify it is a statement accepted without proof. Now suppose someone insists that every accepted statement must itself be proved first. Each of those proofs has a first line too, which needs another accepted statement, and so on. A chain of demands that never ends never gets started, because nothing in it is ever established. A chain that comes back to a statement it has already used is a circle, and a circle assumes the very thing it is meant to prove. Neither proves anything. So every proof begins with statements accepted on their own, and those are the axioms.
Why a world that breaks a statement blocks every proof of it. A valid step has this property: in any world where its reasons are true, its conclusion is true. Suppose every axiom is true in a world. The first step of a proof uses only axioms, so its conclusion is true there. The next step uses axioms and that conclusion, so its conclusion is true there, and so on down the page. Every line of the proof is true in the world, including the last. So a statement that is false in that world cannot be the last line of any proof from those axioms.
The same argument explains why a usable set of axioms cannot contradict itself. If the axioms proved a statement and also its opposite, both would be true in any world where the axioms hold. One of the two is false in every world, so no world can hold the axioms.
What a definition does
Nothing here needs proving in the way a theorem does, but it is worth seeing why unfolding and folding are both allowed.
A definition declares that the word W is shorthand for the condition C. “Is even” is shorthand for “equals 2k for some integer k”, the way “Mon” is shorthand for “Monday”. A shorthand and the phrase it stands for can be swapped in any sentence without changing whether the sentence is true. Swapping the short form for the long form is unfolding. Swapping the long form for the short one is folding. Both moves come from the single fact that W and C were declared to mean the same thing.
It also explains why a definition cannot be proved or disproved. It makes no claim about anything. It fixes how a word will be used, and everything afterwards is a consequence of that choice.
Theorem, lemma and corollary
Why the words add no power. Suppose a proof of a theorem has a line whose reason is Lemma L. The proof of L is a finite list of lines, each justified by axioms, definitions or statements proved before L. Take that list and write it into the proof, at the place where the citation of L stood. Each of its lines is justified as before, and its last line is L, which is exactly what the original line cited. So the new proof reaches the same conclusion without mentioning the lemma.
The lemma’s proof may itself cite earlier results, so do this for every statement of the text in the order the text proves them. The first one is proved from axioms and definitions alone. Each later one cites only statements proved before it, and each of those already has a proof that cites nothing but axioms and definitions, so pasting those proofs in gives the later one such a proof too. By the end, every statement, the theorems included, has a proof with no lemma or corollary cited anywhere in it.
Why there are no circles. Suppose a lemma L were used to prove a theorem T, and T were used to prove L. Then neither is proved before the other, and the pasting above never ends: replacing L brings in T, replacing T brings in L, and so on. This is the circle from the argument that the reasons must stop, and a circle proves nothing.
Why a flaw travels upward. A line in a proof may use only lines above it. If one line has no valid reason, every line that uses it is left unsupported too, and so on up the page. A statement that was proved without that line never used it, so it keeps its proof. That is why the damage spreads toward the statements resting on the flaw and never away from it.
Worked examples
Examples 1 to 6 work inside small axiom systems. Examples 7 to 12 test, unfold and fold definitions. Examples 13 to 19 build theorems, lemmas and corollaries and show each doing its job.
How many lines are in the three-point world?
Use Axioms 1, 2 and 3. Show that there are exactly three lines, and name them.
SolutionBy Axiom 1 the points are three in number. Name them P, Q and R.
Axiom 2 says P and Q lie on exactly one line. Axiom 3 says that line has exactly two points, so it holds P and Q and nothing else. Write it \{P, Q\}. The same two axioms give a line \{P, R\} and a line \{Q, R\}. These three are different lines, because each holds a different pair of points.
There is no fourth line. Any line has exactly two points (Axiom 3), and two points out of P, Q, R are one of those three pairs. Axiom 2 allows only one line through a given pair, so that line is one of the three already named.
So the world has exactly three lines, \{P, Q\}, \{P, R\} and \{Q, R\}, which is what Figure 1 shows. Each point lies on exactly two of them.
How many lines are in a four-point world?
Change Axiom 1 to “There are exactly four points”, and keep Axioms 2 and 3. Find the number of lines, and the number of lines through each point.
SolutionName the points A, B, C and D.
Each line has exactly two points (Axiom 3), so a line names a pair of points. Each pair of points lies on exactly one line (Axiom 2), so a pair names one line. Lines and pairs match one for one, which means the number of lines is the number of pairs.
The number of ways to choose two points from four is
so there are six lines: \{A, B\}, \{A, C\}, \{A, D\}, \{B, C\}, \{B, D\} and \{C, D\}.
Point A is paired with each of the other three points once, so it lies on three lines. The same holds for every point. Figure 6 shows all six lines.
The same reasoning works for any number of points n. There are \displaystyle \binom{n}{2} = \frac{n(n-1)}{2} lines and each point is on n - 1 of them. Notice that the two diagonals cross on the page but share no point: the world contains only the four dots, and the crossing is not one of them.
Proving a theorem from one axiom
Keep only Axioms 2 and 3, with no fixed number of points. Show that two different lines have at most one point in common.
SolutionThis is a statement about every pair of lines, so begin with two different lines and ask what could go wrong. Suppose the lines \ell and m had two different points, X and Y, in common.
Then X and Y both lie on \ell, and both lie on m. But Axiom 2 says two different points lie on exactly one line. Here are two different lines through the same two points, and that contradicts the axiom.
So the supposition was impossible. Two different lines cannot share two points, which means they share one point or none. \blacksquare
The same argument proves the theorem in Euclid’s plane, where the starting axiom is “through two different points there is exactly one line”. The theorem is one step from the axiom, and most theorems are longer chains of steps like this one.
A fact you already know, proved from rules
“Anything times zero is zero” is learned so early that it feels like a rule of its own. It is not. Take these as the starting rules for numbers, along with the agreement that the order and the grouping of adding and of multiplying do not matter:
- R1. a + 0 = a.
- R2. a(b + c) = ab + ac.
- R3. Every number a has an opposite, -a, with a + (-a) = 0.
Prove that a \cdot 0 = 0 for every number a.
SolutionWrite z for the number a \cdot 0. Rewrite the 0 as 0 + 0, which R1 allows because 0 + 0 = 0:
R2 spreads the multiplication over the sum:
Now add the opposite of z, which exists by R3, to both sides. Adding the same thing to two equal quantities keeps them equal:
The left side is 0 by R3. On the right, regroup the sum, then use R3, then R1:
So 0 = z, which says a \cdot 0 = 0. Because the order of multiplying does not matter, 0 \cdot a = 0 too. \blacksquare
So “anything times zero is zero” is a theorem, and R1 to R3 are what it rests on. See properties of real numbers for where those rules come from.
Can the axioms settle it?
Keep Axioms 2 and 3 and drop Axiom 1. Consider the statement
Statement S. Through every point not on a given line there is exactly one line that has no point in common with the given line.
Is S provable, refutable, or undecided?
SolutionThe Results section gives the method: look for worlds that obey the axioms and see whether the statement holds in them.
In the three-point world of Figure 1, take the line \{P, Q\} and the point R off it. The lines through R are \{P, R\} and \{Q, R\}. The first meets \{P, Q\} at P and the second meets it at Q. No line through R avoids \{P, Q\}, so the statement is false in this world.
In the four-point world of Figure 6, take the line \{A, B\} and the point C off it. The lines through C are \{C, A\}, \{C, B\} and \{C, D\}. The first two meet \{A, B\} at A and B. The third, \{C, D\}, shares no point with \{A, B\}. So exactly one line through C misses \{A, B\}. The same thing happens whichever line and point you choose. Call the line \{X, Y\}, the point off it Z, and the fourth point W. The lines \{Z, X\} and \{Z, Y\} meet \{X, Y\}, and \{Z, W\} does not. So the statement is true in this world.
Both worlds obey Axioms 2 and 3. The statement is false in one and true in the other, so no proof of it and no proof of its opposite can exist. It is undecided by Axioms 2 and 3.
This is how the question of Euclid’s parallel postulate, a statement like the one above for the ordinary plane, was finally settled. For two thousand years people tried to prove it from his other starting statements. In the nineteenth century worlds were built in which the others hold and it fails, and that showed no such proof can exist. See the parallel postulate.
Is Axiom 3 needed?
Suppose a system has only Axiom 1 and Axiom 2. Can Axiom 3, “every line has exactly two points”, be proved from them?
SolutionTo show that it cannot, build a world where Axioms 1 and 2 hold and Axiom 3 fails.
Take three points P, Q, R and a single line holding all three.
Axiom 1 holds: there are three points. Axiom 2 holds: any two of the points lie on the one line, and there is no other line for them to lie on, so they lie on exactly one.
Axiom 3 fails: the only line has three points, not two.
By the result “to show that a statement cannot be proved, build a world” in the Results section, no proof of Axiom 3 from Axioms 1 and 2 can exist. Axiom 3 is not a consequence of the other two, and it earns its place in the list.
Testing numbers against a definition
Use the definition of even to decide whether each integer is even: 18, -6, 0, 7.
SolutionFor each number n, ask whether some integer k makes n = 2k true.
18 = 2 \cdot 9, and 9 is an integer, so 18 is even.
-6 = 2 \cdot (-3), and -3 is an integer, so -6 is even. The definition says “some integer k”, not “some positive integer”, so negative values of k are allowed.
0 = 2 \cdot 0, and 0 is an integer, so 0 is even. Nothing about pairs or neighbors was needed, only the test.
For 7, suppose 7 = 2k. Dividing by 2 forces k = 3.5, which is not an integer, and the equation leaves no other choice for k. So no integer works, and 7 is not even. A “no” has to be this complete: showing that one choice of k fails proves nothing, but showing that the equation leaves only a non-integer proves the number fails.
A word from everyday speech
Is a square a rectangle? Use the definition: a rectangle is a quadrilateral with four right angles.
The test asks two things of a shape: is it a quadrilateral, and are all four of its angles right angles.
A square has four sides, so it is a quadrilateral. Its four angles are right angles. It passes both parts, so by folding, a square is a rectangle.
In everyday talk “rectangle” often suggests a shape longer than it is wide. That suggestion was never part of the definition, so it cannot be used. The test contains nothing about side lengths, and the square meets it exactly as the longer rectangle does.
The slanted parallelogram in Figure 7 fails the second part: its angles are 70^\circ and 110^\circ, so they are not right angles.
Unfolding to prove something
Prove that if m and n are even integers, then m + n is even.
SolutionUnfold what is given. Since m is even, m = 2a for some integer a. Since n is even, n = 2b for some integer b. The two unfoldings get different letters, because the definition promises only that such integers exist, and they need not be equal.
Now do ordinary algebra. Adding the two equations,
where the last step takes out a common factor of 2.
Fold. The sum a + b is an integer, because a sum of integers is an integer. So m + n equals 2 times an integer, which is exactly what the definition of even asks for. Hence m + n is even.
Folding to show a property
Show that for every integer n, the number 6n + 4 is even.
SolutionTo fold, we must produce an integer k with 6n + 4 = 2k. Factor out a 2 and see what is left:
So take k = 3n + 2. It is an integer, because n is an integer and integers are closed under multiplication and addition. Then 6n + 4 = 2k, so 6n + 4 is even.
Check. For n = 5, 6n + 4 = 34 and k = 3 \cdot 5 + 2 = 17, and 34 = 2 \cdot 17. For n = -2, 6n + 4 = -8 and k = 3(-2) + 2 = -4, and -8 = 2(-4).
When two books define a word differently
Some books define a trapezoid as a quadrilateral with at least one pair of parallel sides. Others define it as a quadrilateral with exactly one pair. Is every parallelogram a trapezoid?
SolutionThe answer depends on which definition is in force, so work it out under each.
Under the first definition, a parallelogram has two pairs of parallel sides, which is at least one pair. It passes the test, so every parallelogram is a trapezoid.
Under the second definition, a parallelogram has two pairs, not exactly one. It fails the test, so no parallelogram is a trapezoid.
Neither book is wrong, because neither definition is true or false. They are different decisions about one word. What changes is which statements can be proved: under the first definition, every theorem about trapezoids also applies to parallelograms, and under the second it says nothing about them. So a careful writer states which definition is being used, and a careful reader checks.
A word nobody has heard
An integer is called tidy if it equals 4k + 1 for some integer k. First decide whether -3, 21 and 10 are tidy. Then prove that the product of two tidy integers is tidy.
SolutionFor the first part, look for an integer k in each case.
-3 = 4(-1) + 1, so -3 is tidy. 21 = 4 \cdot 5 + 1, so 21 is tidy. For 10, suppose 10 = 4k + 1. Then 4k = 9 and k = \tfrac{9}{4}, which is not an integer, so 10 is not tidy.
For the second part, unfold both given numbers with separate letters. Let m = 4a + 1 and n = 4b + 1, with a and b integers. Multiply:
The number 4ab + a + b is an integer, since a and b are integers. So mn has the form 4k + 1 with k = 4ab + a + b, and folding shows mn is tidy.
Check. 5 = 4 \cdot 1 + 1 and 9 = 4 \cdot 2 + 1 are tidy. With a = 1 and b = 2, the formula gives k = 4 \cdot 1 \cdot 2 + 1 + 2 = 11, and 4 \cdot 11 + 1 = 45 = 5 \cdot 9.
Notice what the proof knew about “tidy”: only its definition. The word was invented a few lines ago, and the argument worked exactly as it would for a word with a long history. A definition is all there is to know about a defined word.
A corollary of the angle sum
Take the angle sum theorem as already proved: the three angles of any triangle add to 180^\circ (see Triangle angle sum, and Proving the triangle angle sum for the proof). (a) Show that the two acute angles of a right triangle add to 90^\circ. (b) One acute angle of a right triangle is twice the other. Find both.
Solution(a) A right triangle has one angle of 90^\circ. Call the other two \alpha and \beta. The angle sum theorem says
Subtracting 90^\circ from both sides gives \alpha + \beta = 90^\circ. \blacksquare
That is the whole proof: one citation and one line of arithmetic. This short, immediate step is what earns the name corollary.
(b) Let the smaller acute angle be x^\circ, so the other is (2x)^\circ. By part (a),
so 3x = 90 and x = 30. The angles are 30^\circ and 60^\circ.
Check. 30 + 60 + 90 = 180. Part (b) cited the corollary of part (a), not the theorem: once proved, a corollary is cited like any other result.
The angles of a quadrilateral
Show that the four angles of a convex quadrilateral add to 360^\circ. Then find the fourth angle of a convex quadrilateral whose other three angles are 85^\circ, 110^\circ and 70^\circ.
Cut the quadrilateral ABCD along the diagonal AC, as in Figure 8. “Convex” means that the segment between any two points of the shape stays inside the shape, so the diagonal runs through the inside and splits the quadrilateral into two triangles, ABC and ACD. The angle sum theorem applies to each triangle:
Now read the corners of the quadrilateral off the figure. The diagonal lies inside the corner at A and inside the corner at C, so each of those corners is cut into two pieces: the corner at A is \angle 1 and \angle 6 together, and the corner at C is \angle 3 and \angle 4 together. The corners at B and D are not cut, so they are \angle 2 and \angle 5. Adding the four corners,
and these are exactly the six angles, so the total is 180^\circ + 180^\circ = 360^\circ. \blacksquare
For the numbers, the fourth angle is whatever is needed to reach 360^\circ:
The fourth angle is 95^\circ.
This is a corollary of the angle sum theorem, but a bigger step than Example 13: it needed a cut and a look at where the pieces go. Whether a step this size is called a corollary or a theorem is the author’s judgment.
A corollary used as a lemma
Show that the five angles of a convex pentagon add to 540^\circ. Then find the fifth angle of a convex pentagon whose other four angles are 100^\circ, 110^\circ, 120^\circ and 95^\circ, and the size of each angle of a regular pentagon.
SolutionLet the pentagon be ABCDE. Cut off the triangle ABC along the diagonal AC. What remains is the quadrilateral ACDE. It is convex, because any four corners of a convex polygon, taken in order, form a convex polygon.
The angles of ABCDE are made of the angles of the two pieces. The corner at A of the pentagon is the triangle’s angle at A together with the quadrilateral’s angle at A, because the diagonal AC lies inside that corner. The same is true at C. The corners at B, D and E belong to one piece each. So the five angles of the pentagon are exactly the three angles of triangle ABC and the four angles of quadrilateral ACDE:
Here the triangle uses the angle sum theorem, and the quadrilateral uses Example 14. The result of Example 14 is now a lemma: it was proved earlier and is used here as a step. \blacksquare
For the missing angle x,
In a regular pentagon the five angles are equal, so each is 540^\circ \div 5 = 108^\circ.
The same cut takes a convex hexagon to a triangle and a pentagon, and so on to any number of sides. Because that step repeats, the general result is usually stated as a theorem in its own right; see Angle sums of polygons.
Proving a lemma
Prove the lemma from the Definition section: every integer square leaves remainder 0 or 1 when divided by 4.
Solution“Leaves remainder r when divided by 4” means the number equals 4q + r for some integer q, with r one of 0, 1, 2, 3. So the goal is to write every square in one of the forms 4q or 4q + 1.
Every integer is even or odd (see even and odd numbers), and unfolding the definitions of even (n = 2m) and odd (n = 2m + 1, as in Exercise 18) gives two cases.
Case 1: n is even. Then n = 2m for some integer m, and
Since m^2 is an integer, n^2 leaves remainder 0.
Case 2: n is odd. Then n = 2m + 1 for some integer m, and expanding the square,
Since m^2 + m is an integer, n^2 leaves remainder 1. \blacksquare
These two cases cover every integer, so every square leaves remainder 0 or 1.
Check. 6^2 = 36 = 4 \cdot 9 + 0, and 7^2 = 49 = 4 \cdot 12 + 1.
A theorem that stands on a lemma
Prove the theorem: if x and y are integers, then x^2 + y^2 does not leave remainder 3 when divided by 4.
SolutionBy the lemma of Example 16, x^2 leaves remainder 0 or 1, and so does y^2. Write x^2 = 4a + r and y^2 = 4b + s, where a and b are integers and each of r and s is 0 or 1. Then
Only four pairs (r, s) are possible, and the table lists them.
| r, from x^2 | s, from y^2 | r + s |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 2 |
In every row r + s is 0, 1 or 2. That is a number from 0 to 3, so it is the remainder of x^2 + y^2 on division by 4 (the remainder is the only such number that works, see There exists, and there is exactly one). It is never 3. \blacksquare
What the lemma bought. Without the lemma, the proof would have to handle x and y directly. There are four cases, x and y each even or odd. Writing x = 2a or 2a + 1 and y = 2b or 2b + 1, the sums are
and the case of x odd and y even is the second line with the letters swapped. So the same proof exists with no lemma in it, as the Results section promised, but it is longer and repeats the squaring of 2m + 1 in three of the four cases. Proved once as a lemma, that squaring is done once.
What a corollary says and does not say
(a) Show that 2027 is not a sum of two integer squares. (b) The theorem rules out remainder 3. Does it say that every number with another remainder is a sum of two squares?
Solution(a) Divide 2027 by 4: 2027 = 4 \cdot 506 + 3, so it leaves remainder 3. By the theorem of Example 17 it is not a sum of two squares, which is the corollary: x^2 + y^2 = 2027 has no integer solution. This took one citation and one division. No search through pairs (x, y) was needed.
(b) No. The theorem says what cannot happen and is silent about the other remainders. The number 6 = 4 \cdot 1 + 2 leaves remainder 2, and it is not a sum of two squares: the squares that are at most 6 are 0, 1 and 4, and 6 - 0 = 6, 6 - 1 = 5 and 6 - 4 = 2 are not squares. On the other side, 2029 = 4 \cdot 507 + 1 leaves remainder 1, and 2029 = 45^2 + 2^2, since 2025 + 4 = 2029. So a remainder other than 3 leaves the question open, and the corollary about 2027 cannot be turned into one about 2029. A theorem of the form “if P then Q” does not give back “if Q then P”; see Converses and biconditional definitions.
The handshake lemma and its corollary
A graph here is a set of dots, called vertices, with some pairs joined by a segment, called an edge. No pair is joined twice and no dot is joined to itself. The degree of a vertex is the number of edges that end at it. The handshake lemma says that the degrees add up to twice the number of edges. (a) Prove it. (b) Prove the corollary that the number of vertices with odd degree is even. (c) In a class of 25 students, friendship is mutual: if one student counts another as a friend, the other counts the first. Can every student have exactly 3 friends in the class?
Solution(a) Count the ends of edges. Every edge has two ends, one at each of its two vertices, so if there are E edges there are 2E ends in all. The degree of a vertex is the number of ends at it, so adding the degrees counts every end once. The sum of the degrees is 2E. \blacksquare
(b) Let S be the sum of all the degrees. By the lemma, S = 2E, which is even. Split S into two parts: the degrees of the even-degree vertices, and the degrees of the odd-degree vertices. The first part is a sum of even numbers, so it is even. Then the second part is S minus an even number, so it is even too.
Suppose there are k odd-degree vertices, with degrees 2c_1 + 1, \dots, 2c_k + 1. Their sum is
which is even exactly when k is even. The second part is even, so k is even. \blacksquare
As a story: at any party, the number of guests who shook an odd number of hands is even.
(c) Make each student a vertex and each friendship an edge. If every student had exactly 3 friends, all 25 vertices would have odd degree. But 25 is odd, and the corollary says the number of odd-degree vertices is even. So it is impossible.
Check. The lemma says the same: the degrees would add to 25 \cdot 3 = 75, which is odd, yet a sum of degrees must equal 2E, an even number.
Boundaries
What an axiom is
An axiom is accepted, not announced as true of the real world. The axioms of the plane say what must be so if something is to count as a Euclidean plane. Whether a chalkboard, or a soccer field, or space itself follows them is a separate question, answered by measuring and not by proving.
A set of axioms must not contradict itself. A system that proves a statement and also its opposite is useless, since in ordinary logic it then proves every statement whatever. Showing that a set of axioms is free of contradiction usually means exhibiting a world in which all of them hold, which is what Figure 1 does for the three-point axioms.
Accepted without proof is not accepted without reason. Axioms are chosen because they are simple, few, and strong enough to build what is wanted. But they can be replaced. Swap one, and you are describing a different world, not correcting a mistake.
The axioms do not always settle every question. Example 5 found a statement that Axioms 2 and 3 leave open. The same thing happens at a far larger scale. A result of Gödel, stated here without proof, says this: take any system whose axioms do not contradict each other, whose axioms can be listed by a fixed rule (a computer program could print them all), and which can prove the basic facts about adding and multiplying whole numbers. Then there is a statement about whole numbers that the system can neither prove nor refute. (Gödel proved it in 1931 with a slightly stronger assumption in place of “do not contradict each other”; Rosser removed that extra assumption in 1936.)
An axiom is not a conjecture. A conjecture is a statement nobody has proved yet, and it is hoped that a proof exists. An axiom is a statement nobody intends to prove.
What a definition does
A definition says which things a word names. It does not say there are any. The sentence “call an integer huge if no integer is larger than it” is a perfectly good definition, with a clear test. But for any integer n, the integer n + 1 is larger, so nothing passes the test. Whenever it matters that a defined thing exists, that has to be shown separately (see Existence proof).
A definition says only what its condition says. The definition of rectangle asks for four right angles. That the diagonals of a rectangle are equal is true, but it is not in the definition, so it cannot be used until it has been proved. The same goes the other way: a feeling about what a word suggests, such as “a rectangle is longer than it is wide”, is not part of it.
A definition covers only the kind of thing it names. Even is defined for integers. The number 2.5 is not an integer, so the definition does not make it even, and the usual convention is that only integers are called even or odd.
It cannot refer to itself. Suppose a word were defined by “an integer is bonky if it is one more than a bonky integer”. To test whether 5 is bonky you would ask whether 4 is, then 3, and so on without end. The test never reaches a place where the question is answered. Every word in a definition has to be settled before the definition begins, and at the very bottom some terms are left undefined on purpose, exactly as the points and lines of an axiom system are.
Two definitions of one word may both be in use. When each condition implies the other, they name exactly the same things and either may be chosen, the other becoming a theorem. When they do not, as with the two trapezoids of Example 11, the word means different things in different books, and nothing can be carried between them unchecked.
Theorem, lemma and corollary
The words describe a role. They do not measure difficulty or importance. A lemma may be much harder than the theorem it serves, and a name can stay after the role has changed. Zorn’s lemma is still called a lemma, yet in standard set theory it is known to be equivalent to the axiom of choice, which makes it a cornerstone rather than a small helper.
The labels are a judgment, not a test. The word even has an exact condition that any integer either meets or fails. “Main result”, “helper” and “quick consequence” have no such condition, so a result can be a theorem in one book, where it is the point of a chapter, and a lemma in another, where it is a step toward something bigger. How big a step a corollary may be is a judgment of the same kind, as Examples 13 and 14 show. What is not a matter of judgment is whether the statement is proved.
Names can outlive their meaning. The statement called Fermat’s Last Theorem was named that long before anyone had a proof of it, and it was proved only in the 1990s. So a name, taken alone, does not tell you whether a proof exists. Inside a text, a statement labeled a theorem is a promise that a proof follows.
Only proved statements get these labels. A conjecture, an axiom and a definition, all met above, have no proof, so none of them is a theorem, a lemma or a corollary.
A corollary reaches only as far as its proof. It follows from its parent, perhaps with facts already proved, and can claim nothing those do not support. In particular the converse of a theorem is not a corollary of it: it is a separate statement that needs its own proof, and it may be false, as Example 18 shows.
Where this goes next
Once you know where a system starts and what its words mean, you can read how a text is built. What may be assumed and what must be shown looks at what a proof may start from, and the answer always includes the axioms and the definitions of its words. A direct proof is unfold, reason, fold, chained from the starting statements to a conclusion, and what to do when that will not work is proof by contradiction, which Example 3 used. A long proof is usually a theorem standing on lemmas, each proved by one of those methods.
Proving statements about even and odd numbers and Proving statements about divisibility are the definition moves applied to two particular definitions, and the tidy example is what they look like with the name changed. The counting view of even numbers is even and odd numbers, and factors and multiples holds the definition that most of number theory starts from.
For geometry in particular, converses and biconditional definitions sorts every statement in a proof into what was agreed, what was named, and what was shown, and explains why a definition reads both ways. The axioms of geometry start from the words left undefined, which the lesson on undefined terms describes, and the most famous of them, the one that Example 5 echoes, is the parallel postulate. Two places where the exact definition matters are trapezoids and rectangles. The cut in Examples 14 and 15 is the method of Angle sums of polygons, and the lemma style of the last example continues in degree and the handshake lemma.
Exercises
Reading an axiom system
Use the three axioms of the three-point world for Exercises 1 to 4.
- How many lines pass through the point P?
- How many points do two different lines of this world have in common?
- Is “some line contains all three points” provable, refutable or undecided? Explain.
- Is “the point P is to the left of the point Q” provable, refutable or undecided by these axioms? Explain.
Counting in other worlds
For Exercises 5 to 9 keep Axioms 2 and 3 but change the number of points.
- A world has exactly 4 points. How many lines are there, and how many lines pass through each point?
- A world has exactly 5 points. How many lines are there, and how many lines pass through each point?
- A world has exactly 12 points. How many lines are there?
- A world has exactly 45 lines. How many points does it have?
- Can a world of this kind have exactly 20 lines? Explain.
Axioms for numbers
Use R1 to R3 from Example 4, together with R4: 1 \cdot a = a for every number a.
- Use R2 to compute 7 \cdot 13 + 7 \cdot 87 without multiplying 7 by 13.
- Write 97 as 100 + (-3) and use R2 to compute 6 \cdot 97.
- Show that a + (-1) \cdot a = 0 for every number a.
Worlds and independence
- A world has four points A, B, C, D, and its lines are the four sets \{A, B, C\}, \{A, B, D\}, \{A, C, D\} and \{B, C, D\}. How many lines hold both A and B? Which of Axioms 2 and 3 fail in this world?
- A world has three points P, Q, R and exactly two lines, \{P, Q\} and \{P, R\}. Which of Axioms 1, 2 and 3 hold? What does this show about Axiom 2?
- On a globe, let a point be a place on the surface and a line be a great circle, such as the equator or the circle through both poles along the 0^\circ and 180^\circ meridians. Does every pair of points lie on exactly one line? Are there two lines that never meet?
Testing against a definition
- Use the definition of even to decide whether each integer is even: 26, -14, 91, -1.
- An integer is a multiple of 5 if it equals 5k for some integer k. Decide whether each is a multiple of 5: 35, -20, 0, 52.
- An integer is odd if it equals 2k + 1 for some integer k. Decide whether each is odd: 15, -3, 0, 100.
- A positive integer is a perfect square if it equals k^2 for some positive integer k. Decide whether each is a perfect square under this definition: 49, 50, 1, 0.
Geometry words
- A rhombus is a quadrilateral whose four sides all have the same length. Use this definition to decide whether a square is a rhombus, and whether every rhombus is a square. Give an example that settles the second question.
- A quadrilateral has three right angles. The four angles of any quadrilateral add to 360^\circ. Find the fourth angle, and say whether the shape is a rectangle under the definition in Example 8.
- A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Using the definition of trapezoid with “at least one pair of parallel sides”, decide whether every parallelogram is a trapezoid, and whether every trapezoid is a parallelogram. For the second question, settle it with a quadrilateral whose corners you give as coordinates.
Unfolding and folding
- Let n be an even integer. Show that n + 6 is even.
- Show that for every integer n, the number 10n - 8 is even.
- Show that for every integer n, the number 4n^2 + 2n is even.
- Let m and n be even integers. Show that m - n is even.
- Let n be an even integer. Show that 3n is even.
- Let m be an even integer and n any integer. Show that mn is even.
Words made up for the occasion
For Exercises 29 to 32, an integer is a zorp if it equals 5k + 2 for some integer k.
- Decide whether each is a zorp: 17, -3, 25, 2.
- Show that 5n + 7 is a zorp for every integer n.
- Show that the sum of a zorp and a multiple of 5 is a zorp.
- Show that the product of two zorps equals 5k + 4 for some integer k, and that no product of two zorps is a zorp.
Corollaries of the angle sum
- One acute angle of a right triangle is 28^\circ. Find the other.
- The angles of a triangle are x^\circ, (x + 10)^\circ and (x + 20)^\circ. Find all three.
- Three angles of a convex quadrilateral are 88^\circ, 112^\circ and 67^\circ. Find the fourth.
- The four angles of a convex quadrilateral are x^\circ, 2x^\circ, 3x^\circ and 4x^\circ. Find all four.
- Four angles of a convex pentagon are 110^\circ, 120^\circ, 100^\circ and 130^\circ. Find the fifth.
- Cut a triangle off a convex pentagon in the way of Example 15 to find the sum of the angles of a convex hexagon, and then each angle of a regular hexagon.
- A triangle has angles of 50^\circ at A and 65^\circ at B. Side BC is extended past C, and the angle between the extension and side CA is called the exterior angle at C. Find it.
Squares and remainders
- What remainder does 37^2 leave when divided by 4?
- The theorem of Example 17 rules out some of the numbers 2027, 2029, 2031 and 2035 as sums of two squares. Which ones? Then write the remaining number as a sum of two squares.
- What remainders can the cube n^3 of an integer leave when divided by 4? Write n = 4m + r with r one of 0, 1, 2, 3.
- Show that the square of an integer leaves remainder 0 or 1 when divided by 3. Write n = 3m + r with r one of 0, 1, 2.
- Use Exercise 43 to prove: if x and y are integers and x^2 + y^2 is divisible by 3, then both x and y are divisible by 3.
- Use Exercise 44 twice to find every pair of non-negative integers x \le y with x^2 + y^2 = 2025.
Build a lemma, then a theorem
- Show that the square of an integer leaves remainder 0, 1 or 4 when divided by 8. Use n = 4m + r with r one of 0, 1, 2, 3.
- Use Exercise 46 to show that a sum of three squares of integers never leaves remainder 7 when divided by 8. Then write down two numbers that cannot be written as a sum of three squares.
- The theorem of Example 17 rules out remainder 3 when dividing by 4. Give a number that leaves remainder 1 when divided by 4 and is not a sum of two squares.
The handshake
- Explain why no graph has five vertices with degrees 3, 3, 3, 2 and 2.
- The degrees 3, 3, 3, 1 pass the test of the corollary: there are four odd degrees. Show that even so, no graph has four vertices with these degrees.
Reasoning about axioms
- Explain why a proof cannot consist entirely of statements that each have a proof inside the same proof.
- A friend says an axiom is “a statement that is obviously true”. Give two reasons that this is not what makes a statement an axiom.
- Suppose the axioms of a system prove a statement and also its opposite. What does this say about the system?
- A statement is proved from Axioms 1 and 2. A student then finds a world where both axioms hold and the statement is false. What must be true?
- One axiom of a system can be proved from the others. Is something wrong?
Reasoning about definitions
- A friend says zero cannot be even “because there is nothing to pair up”. Answer the friend using the definition, and say why the feeling about pairs does not settle the matter.
- Someone asks you to “prove that the definition of even is correct”. Explain why that cannot be done, and name two things it does make sense to ask about a definition.
- A student writes: “Let m and n be even. Then m = 2k and n = 2k, so m + n = 4k = 2(2k), which is even.” What has this argument actually covered, and how is it repaired?
- Write a circular definition of a made-up word, and explain what goes wrong when you try to decide whether a particular number meets it.
- Call an integer huge if no integer is larger than it. Is that a legitimate definition? Is any integer huge?
- One book defines a rectangle as a parallelogram with four right angles. Another defines it as a quadrilateral with four right angles. Show that both books name exactly the same shapes. You may use this fact: if two lines are cut by a third line, and the two interior angles on one side add to 180^\circ, the two lines are parallel.
Reasoning about theorems, lemmas and corollaries
- An author proves that every square leaves remainder 0 or 1 when divided by 4, uses that in the proof of one later result, and never mentions it again. Which of the three words suits it best, and why?
- A friend says that “a lemma is a theorem that is not quite as true”. Explain what is wrong with that, and what the three words do distinguish.
- A lemma is written to help prove a theorem. Explain why the proof of the lemma cannot use that theorem.
- A result is called a lemma in one book and a theorem in another. Can both books be right?
- In Example 14 the diagonal AC was assumed to lie inside the quadrilateral. A dart-shaped quadrilateral has one angle greater than 180^\circ, at a corner C, and its diagonal BD lies outside it. Which diagonal can you use to repeat the argument, and why does the argument then still give 360^\circ?
- Prove from the angle sum theorem that an exterior angle of a triangle equals the sum of the two interior angles at the other two corners.
Answers
- Two. Each point lies on a line with each of the other two points, and there is one line per pair, so P is on \{P, Q\} and \{P, R\} and on no other line.
- One point. Two different lines are two different pairs from three points, and two different pairs out of three points share exactly one point. For example, \{P, Q\} and \{P, R\} share P.
- Refutable. Axiom 3 says every line has exactly two points, so a line with three points contradicts an axiom, and the opposite statement is provable.
- Undecided. The axioms mention only points, lines and which points are on which lines. They say nothing about left or right, so neither the statement nor its opposite can be proved. The left-right order in Figure 1 belongs to the picture, not to the world.
- Six lines, and three through each point. Lines and pairs match one for one, and \binom{4}{2} = 6. Each point is paired with the other 3 points.
- Ten lines, and four through each point. \displaystyle \binom{5}{2} = \frac{5 \cdot 4}{2} = 10, and each point is paired with the other 4 points.
- 66 lines. \displaystyle \binom{12}{2} = \frac{12 \cdot 11}{2} = 66.
- 10 points. The number of lines with n points is \displaystyle \frac{n(n-1)}{2}, so n(n-1) = 90. Since 10 \cdot 9 = 90, n = 10. (The other solution, n = -9, is not a number of points.)
- No. With 6 points there are 15 lines and with 7 points there are 21, and the number of lines grows at every step, so 20 is skipped. No number of points gives 20.
- 700. By R2 read backwards, 7 \cdot 13 + 7 \cdot 87 = 7(13 + 87) = 7 \cdot 100 = 700.
- 582. By R2, 6(100 + (-3)) = 6 \cdot 100 + 6 \cdot (-3), and that is 600 - 18 = 582.
- Use R4 to write a = 1 \cdot a, so that a + (-1) \cdot a = 1 \cdot a + (-1) \cdot a. R2 read backwards turns the right side into (1 + (-1)) \cdot a, which is 0 \cdot a by R3, and 0 \cdot a = 0 by Example 4. This says that (-1) \cdot a does the job of the opposite of a.
- Two lines hold both A and B, namely \{A, B, C\} and \{A, B, D\}, so Axiom 2 fails: it asks for exactly one line through two different points. Axiom 3 fails as well, since each line has three points and not two.
- Axioms 1 and 3 hold: there are three points and each line has two. Axiom 2 fails, because Q and R lie on no line at all. So Axiom 2 is not a consequence of Axioms 1 and 3 together, since this world keeps both and breaks it.
- No to the first question. Two points at opposite ends of a diameter, such as the north and south poles, lie on infinitely many lines, namely every great circle through both poles. Any other two points lie on exactly one great circle. No to the second: any two great circles meet, in two opposite points. So “exactly one line” fails for some pairs, and no two lines miss each other.
- 26 and -14 are even; 91 and -1 are not. 26 = 2 \cdot 13 and -14 = 2 \cdot (-7). For 91 = 2k the equation forces k = 45.5, and for -1 = 2k it forces k = -0.5, neither an integer.
- 35, -20 and 0 are multiples of 5; 52 is not. 35 = 5 \cdot 7, -20 = 5 \cdot (-4) and 0 = 5 \cdot 0. For 52 = 5k the equation forces k = 10.4, which is not an integer.
- 15 and -3 are odd; 0 and 100 are not. 15 = 2 \cdot 7 + 1 and -3 = 2(-2) + 1. For 0 = 2k + 1 the equation forces k = -0.5, and for 100 = 2k + 1 it forces k = 49.5.
- 49 and 1 are perfect squares; 50 and 0 are not. 49 = 7^2 and 1 = 1^2. For 50 we would need \displaystyle k = \sqrt{50}, which lies between 7 and 8 and is not an integer. And 0 is not a positive integer, so this definition does not name it. A book that wanted 0 included would say “non-negative”.
- A square is a rhombus, but a rhombus need not be a square. A square’s four sides are equal, so it passes the test. The test says nothing about angles, so a rhombus with angles of 60^\circ and 120^\circ (two equilateral triangles placed edge to edge) passes it, yet has no right angles, so it is not a square.
- The fourth angle is 90^\circ, and the shape is a rectangle. 360^\circ - 270^\circ = 90^\circ, so all four angles are right angles, which is the whole of the test, and folding gives a rectangle.
- Every parallelogram is a trapezoid, but not every trapezoid is a parallelogram. A parallelogram has two pairs of parallel sides, so it has at least one pair, and passes the trapezoid test. For the other way, take corners (0,0), (4,0), (3,2), (1,2). The sides from (0,0) to (4,0) and from (1,2) to (3,2) are both horizontal, so they are parallel and the shape is a trapezoid. The other two sides have slopes 2 and -2, so they are not parallel, and the shape is not a parallelogram.
- Unfold: n = 2k for some integer k. Then n + 6 = 2k + 6 = 2(k + 3), and k + 3 is an integer, so n + 6 is even by folding.
- 10n - 8 = 2(5n - 4), and 5n - 4 is an integer because n is, so 10n - 8 is even.
- 4n^2 + 2n = 2(2n^2 + n), and 2n^2 + n is an integer because n is, so the number is even.
- Unfold both: m = 2a and n = 2b for integers a and b. Then m - n = 2a - 2b = 2(a - b), and a - b is an integer, so m - n is even.
- Unfold: n = 2k for some integer k. Then 3n = 6k = 2(3k), and 3k is an integer, so 3n is even.
- Unfold: m = 2a for some integer a. Then mn = 2an = 2(an), and an is an integer because a and n are, so mn is even. Nothing was assumed about n beyond its being an integer.
- 17, -3 and 2 are zorps; 25 is not. 17 = 5 \cdot 3 + 2, -3 = 5(-1) + 2 and 2 = 5 \cdot 0 + 2. For 25 = 5k + 2 the equation forces k = 4.6, which is not an integer.
- 5n + 7 = 5(n + 1) + 2, and n + 1 is an integer because n is, so 5n + 7 is a zorp.
- Unfold: the zorp is 5a + 2 and the multiple of 5 is 5b, for integers a and b. Their sum is 5a + 2 + 5b = 5(a + b) + 2, and a + b is an integer, so the sum is a zorp.
- Unfold with separate letters: the zorps are 5a + 2 and 5b + 2. Their product is (5a + 2)(5b + 2) = 25ab + 10a + 10b + 4, which equals 5(5ab + 2a + 2b) + 4. So it is 5k + 4 with k = 5ab + 2a + 2b an integer. If it were a zorp it would also equal 5j + 2 for some integer j, giving 5k + 4 = 5j + 2 and so 5(k - j) = -2. That would make -2 a multiple of 5, but -2 = 5m forces m = -0.4, which is not an integer. So no product of two zorps is a zorp.
- 62^\circ. By the corollary of Example 13 the acute angles add to 90^\circ, so the other is 90^\circ - 28^\circ = 62^\circ.
- 50^\circ, 60^\circ and 70^\circ. The angle sum gives x + (x + 10) + (x + 20) = 180, so 3x + 30 = 180 and x = 50.
- 93^\circ. The four angles add to 360^\circ, and 88 + 112 + 67 = 267, so the fourth is 360 - 267 = 93.
- 36^\circ, 72^\circ, 108^\circ and 144^\circ. The four angles add to 360^\circ, so 10x = 360 and x = 36. Each is less than 180^\circ, as a convex shape needs.
- 80^\circ. The five angles add to 540^\circ, and 110 + 120 + 100 + 130 = 460, so the fifth is 540 - 460 = 80.
- The hexagon’s angles add to 720^\circ, and each angle of a regular hexagon is 120^\circ. Cut a triangle off along a diagonal joining two corners with one corner between them. The rest is a convex pentagon, so the total is 180^\circ + 540^\circ = 720^\circ. A regular hexagon has six equal angles, so each is 720^\circ \div 6 = 120^\circ.
- 115^\circ. The angle at C inside the triangle is 180^\circ - 50^\circ - 65^\circ = 65^\circ by the angle sum. The exterior angle and the interior angle at C fill a straight line, so the exterior angle is 180^\circ - 65^\circ = 115^\circ. This equals 50^\circ + 65^\circ.
- 1. The number 37 is odd, so the second case of the lemma in Example 16 applies, and 37^2 = 1369 = 4 \cdot 342 + 1.
- The theorem rules out 2027, 2031 and 2035, and the remaining number is 2029 = 45^2 + 2^2. Each of 2027 = 4 \cdot 506 + 3, 2031 = 4 \cdot 507 + 3 and 2035 = 4 \cdot 508 + 3 leaves remainder 3. The number 2029 = 4 \cdot 507 + 1 leaves remainder 1, so it is not ruled out, and 45^2 + 2^2 = 2025 + 4 = 2029.
- The remainders 0, 1 and 3. Expanding, (4m + r)^3 = 64m^3 + 48m^2r + 12mr^2 + r^3, and every term but r^3 is a multiple of 4. So n^3 leaves the same remainder as r^3 does: r = 0, 1, 2, 3 gives r^3 = 0, 1, 8, 27, which leave remainders 0, 1, 0, 3.
- For n = 3m: n^2 = 9m^2 = 3(3m^2), remainder 0. For n = 3m + 1: n^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1, remainder 1. For n = 3m + 2: n^2 = 9m^2 + 12m + 4 = 3(3m^2 + 4m + 1) + 1, remainder 1. These cover every integer, so a square leaves remainder 0 or 1.
- By Exercise 43, x^2 and y^2 each leave remainder 0 or 1 on division by 3. The sum x^2 + y^2 is divisible by 3 only if the two remainders add to a multiple of 3: the sums are 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1 and 1 + 1 = 2, and only 0 + 0 works. So x^2 and y^2 are both divisible by 3. By the cases in Exercise 43, an integer that is not divisible by 3 has a square with remainder 1, so x and y must each be divisible by 3.
- (0, 45) and (27, 36). Since 2025 = 3 \cdot 675 is divisible by 3, Exercise 44 says x = 3a and y = 3b, and dividing by 9 gives a^2 + b^2 = 225. This is divisible by 3 again, so a = 3c and b = 3d, and dividing by 9 gives c^2 + d^2 = 25. The pairs with c \le d are (0, 5) and (3, 4), and multiplying back by 9 gives (0, 45) and (27, 36). Check: 27^2 + 36^2 = 729 + 1296 = 2025.
- For n = 4m: n^2 = 16m^2 = 8(2m^2), remainder 0. For n = 4m + 1: n^2 = 16m^2 + 8m + 1 = 8(2m^2 + m) + 1, remainder 1. For n = 4m + 2: n^2 = 16m^2 + 16m + 4 = 8(2m^2 + 2m) + 4, remainder 4. For n = 4m + 3: n^2 = 16m^2 + 24m + 9 = 8(2m^2 + 3m + 1) + 1, remainder 1. So the remainders are 0, 1 and 4.
- By Exercise 46 each of the three squares leaves remainder 0, 1 or 4. Adding three numbers from \{0, 1, 4\} gives one of 0, 1, 2, 3, 4, 5, 6, 8, 9, 12, and none of these leaves remainder 7 after division by 8 (they leave 0, 1, 2, 3, 4, 5, 6, 0, 1, 4). So no sum of three squares leaves remainder 7. For example 31 = 8 \cdot 3 + 7 and 1999 = 8 \cdot 249 + 7 cannot be written as a sum of three squares.
- 21. It leaves remainder 1, since 21 = 4 \cdot 5 + 1. The squares at most 21 are 0, 1, 4, 9, 16, and 21 minus each of them is 21, 20, 17, 12, 5, none of which is a square. So 21 is not a sum of two squares, even though the theorem does not rule it out.
- The degrees add to 3 + 3 + 3 + 2 + 2 = 13, which is odd. By the handshake lemma the sum of the degrees is twice the number of edges, which is even. So no such graph exists. (The corollary gives the same answer: there are three odd degrees, and 3 is odd.)
- Suppose three of the four vertices each have degree 3. A vertex of degree 3 in a graph with four vertices is joined to all three of the others. So each of the three is joined to the fourth vertex, and the fourth vertex has degree at least 3, not 1. So no graph has these degrees. The corollary could not see this, because it only gives a necessary condition.
- Each such statement needs a reason that is itself a statement proved inside the proof, which needs another, and so on. The proof is finite, so the reasons cannot keep going. Either some statement is accepted without a proof inside the proof, or the reasons go in a circle, and a circle assumes what it is proving.
- Two of these. A set of axioms can be changed, and the changed set is still a fair starting point: the four-point and three-point worlds obey different axioms and neither is a mistake. So “obvious” is not the test. Also, statements that sound obvious can be wrong: the rule that every description picks out a set sounds obvious and leads to a contradiction. An axiom is a statement placed at the start on purpose, whatever the reason.
- The axioms contradict one another, and no world satisfies all of them. The system is useless, since every statement whatever can then be proved in ordinary logic.
- There is a mistake: the proof is wrong, or the world does not really obey both axioms. A valid proof from Axioms 1 and 2 ends in a statement that is true in every world where both axioms are true, so it cannot be false in this one.
- Nothing is wrong, but the axiom is redundant. Delete it and everything provable before is still provable, because the deleted axiom can be proved from what is left. Whether to keep it is a matter of taste: a shorter list is neater, a longer one can be easier to use.
- The definition asks for one thing: an integer k with 0 = 2k. The choice k = 0 works, so 0 is even. The picture of pairs is a way of remembering what “even” means, not part of what it was defined to mean, and when a picture and a definition disagree the definition decides, because it is what was agreed. Here they do not truly disagree: zero pairs, with nothing left over.
- A definition is not a claim. It is a decision about how a word will be used, so there is no statement in it that could be true or false. Sensible questions are whether every word in it is already understood (it is not circular), whether anything meets it (it is not empty), and whether it agrees with another definition in use for the same word.
- It covers only the case m = n: using the same letter k for both gives m = 2k = n. The repair is a separate letter for each unfolding. Write m = 2a and n = 2b, so m + n = 2a + 2b = 2(a + b), which covers every pair of even integers.
- One example: “an integer is bonky if it is one more than a bonky integer”. To decide whether 5 is bonky you must decide whether 4 is, which needs 3, and so on without end. The test refers to itself and never reaches a starting point, so it never gives a verdict.
- It is a legitimate definition: every word in it is understood and it gives a clear test. But no integer passes the test, since for any integer n the integer n + 1 is larger. So no integer is huge, and nothing proved “for every huge integer” is about anything.
- Call the book with “parallelogram” Book A and the other Book B. Every shape Book A names has four right angles and is a quadrilateral, so Book B names it too. Conversely, take a quadrilateral ABCD with four right angles. No angle is more than 180^\circ, so the shape is convex and A and D lie on the same side of the line BC. Sides AB and CD are cut by that line, and the interior angles on that side, at B and C, are 90^\circ + 90^\circ = 180^\circ, so AB is parallel to CD. In the same way, sides AD and BC are cut by the line CD, and the angles at C and D add to 180^\circ, so AD is parallel to BC. Both pairs of opposite sides are parallel, so the shape is a parallelogram with four right angles, and Book A names it too. The two books name the same shapes, so each definition makes the other a theorem.
- A lemma. It is there only to serve a later proof and is used once, which is the job of a stepping stone. It is not a main result, and it is not a quick consequence of something just proved.
- A lemma is proved exactly as firmly as a theorem, so it is not less true. The three words name jobs: a theorem is a main result, a lemma is mainly a help for a later proof, and a corollary is a quick consequence of a result just proved.
- If the proof of the lemma used the theorem, and the theorem’s proof used the lemma, each would be assumed while proving the other. That is a circle, and a circle proves nothing. Neither statement would be established from the axioms.
- Yes. The labels record each author’s view of what is a main result and what is a stepping stone, and those can differ. A statement can be the point of one book and a step in another. What must agree is that the statement is proved, and both books prove it.
- Use the diagonal CA, from the corner with the angle greater than 180^\circ. It lies inside the dart, because it starts at the corner that points inward and runs to the opposite corner, while BD lies outside. It cuts the dart into two triangles, ABC and ACD. The corner at C is split into the two angles of the triangles at C, and the corner at A is split into the two angles at A, exactly as before. The six triangle angles still make up the four corners, so the total is again 180^\circ + 180^\circ = 360^\circ.
- Let the triangle be ABC, with side BC extended past C and exterior angle \varepsilon at C. The exterior angle and the interior angle at C fill a straight line, so \varepsilon = 180^\circ - m\angle C. The angle sum theorem gives m\angle A + m\angle B = 180^\circ - m\angle C. The two right sides are equal, so \varepsilon = m\angle A + m\angle B.