Approximating irrational numbers
The square root of 29 is not a whole number, and its decimal never ends. How can you tell how big it is, and where it sits among other numbers?
Where you are 5 in Irrational and real numbers
When you reach for it
A square garden has an area of 90 square meters. Its side is the positive number that squares to 90, which is \displaystyle \sqrt{90}, and 90 is not one of the perfect squares. It is not a fraction either: the square root of a whole number that is not a perfect square is irrational, and its decimal never ends or repeats. You can still say how long the side is to the nearest tenth of a meter, and whether it is longer than some other length.
Reach for this lesson whenever an irrational number has to become a size you can use: to measure with, to mark on a number line, to compare, or to estimate an expression such as \displaystyle 2 + \sqrt{3} or \pi^2. The method needs nothing but multiplication, and unlike a calculator it shows why the digits are right.
This lesson is taught over two days. Part 1 traps a square root between two integers and then narrows the trap to tenths and hundredths (Examples 1 to 4). Part 2 places irrational numbers on the number line, puts them in order, and estimates expressions built from them (Examples 5 to 7).
The procedure
Part 1: trapping a square root
One fact does all the work. For two positive numbers a and b,
Squaring keeps positive numbers in the same order. Here is why. If 0 < a < b, multiply both sides of a < b by the positive number a to get a^2 < ab, and by the positive number b to get ab < b^2. Together, a^2 < ab < b^2. In the other direction, if a were not less than b, then b \le a, and the same argument with the letters swapped gives b^2 \le a^2. So a^2 < b^2 can only happen when a < b.
This turns every question about a square root into a question about squares, and squares you can work out. Is \displaystyle \sqrt{29} bigger than 5? Square both: \displaystyle (\sqrt{29})^2 = 29 and 5^2 = 25. Since 25 < 29, the fact gives \displaystyle 5 < \sqrt{29}.
Trap it between two integers. Find the perfect squares on either side of the number. 29 lies between 25 = 5^2 and 36 = 6^2:
Each of these is a square, \displaystyle (5)^2 < (\sqrt{29})^2 < (6)^2, so the fact undoes all three squares at once and keeps the order:
Figure 1 draws the same thing. The upper line carries the numbers and the lower line their square roots, and each perfect square is joined to its root.
The gaps between perfect squares grow (25 to 36 is 11, but 36 to 49 is 13), so the two lines are not scaled alike, and the place of 29 between 25 and 36 gives only a rough idea of where \displaystyle \sqrt{29} sits between 5 and 6.
Narrow it to tenths. Between 5 and 6 lie the tenths 5.1, 5.2, \ldots, 5.9, and \displaystyle \sqrt{29} lies between two neighboring ones. Squaring finds which. A first guess saves work: 29 is 4 of the 11 steps from 25 to 36, a bit over a third of the way, so try near 5.4.
The first is too big and the second too small, so 5.3^2 < 29 < 5.4^2 and
The guess only decides which squares to try first; a poor guess costs an extra multiplication, never a wrong result.
Round by squaring the halfway point. To give \displaystyle \sqrt{29} to the nearest tenth you need to know which of 5.3 and 5.4 it is nearer, so square the number halfway between them: 5.35^2 = 28.6225. That is less than 29, so \displaystyle 5.35 < \sqrt{29}, and \displaystyle \sqrt{29} is in the upper half of the interval. To the nearest tenth, \displaystyle \sqrt{29} \approx 5.4.
Keep going for more digits. The next digit is the same move one place further: the hundredths between 5.3 and 5.4, as in Example 2. Each round makes the trap ten times shorter. It never closes completely, because no decimal that stops squares to exactly 29 (Exercise 25 asks why), so the result is always written with \approx, never =.
Lengths from areas. A square of area A has a side s with s^2 = A and s > 0, so \displaystyle s = \sqrt{A}. A length written as a multiple of a square root can be put under one root sign too: for a positive number c, \displaystyle (c\sqrt{2})^2 = c^2 \cdot (\sqrt{2})^2 = 2c^2, because a product is squared by squaring each factor. So \displaystyle c\sqrt{2} = \sqrt{2c^2}, and Example 4 shows why that helps.
Part 2: placing, comparing and estimating
Placing a number on the line. Once you know a < x < b, the point x lies between the points a and b. Two moves extend this to numbers built from square roots.
A sum shifts the trap. Adding the same number c to all three parts keeps the order, so
A negative mirrors it. Multiplying by -1 reverses an inequality, so the ends swap:
On the line, -x is the mirror image of x in 0, and the end that was on the right is now on the left.
The number \pi is irrational too, but it is not the square root of any whole number, so trial squaring cannot trap it. Its decimals come from the circle, as in Circumference and Area of a circle: \pi = 3.14159\ldots, so for example 3.14 < \pi < 3.15 and 3.141 < \pi < 3.142.
Comparing two numbers. When both are positive, the larger square belongs to the larger number. When squaring is awkward, as with \pi or a sum of two roots, trap each number and narrow until the two traps no longer overlap.
Estimating an expression. Traps combine. Suppose a < x < b and c < y < d.
- For a sum, add the lower ends and add the upper ends: a + c < x + y < b + d.
- For a difference x - y, the result is smallest when x is small and y is large, so the ends cross over: a - d < x - y < b - c.
- For a product of positive numbers, multiply the lower ends and multiply the upper ends: ac < xy < bd. This is the fact about squaring with two different numbers, proved the same way.
The result is again a trap. If every number in it rounds to the same value, that value is the estimate. If the trap straddles a rounding point, tighten the starting traps by one more decimal place and combine again.
Worked examples
Trapping three square roots between integers
Between which two consecutive integers does each number lie?
(a) \displaystyle \sqrt{67} \qquad (b) \displaystyle \sqrt{190} \qquad (c) \displaystyle \sqrt{700}, and which of the two integers is it nearer?
SolutionIn each case find the perfect squares on either side and undo the squares, which keeps the order because every number involved is positive.
(a) 8^2 = 64 and 9^2 = 81, and 64 < 67 < 81. So \displaystyle 8 < \sqrt{67} < 9.
(b) 13^2 = 169 and 14^2 = 196, and 169 < 190 < 196. So \displaystyle 13 < \sqrt{190} < 14.
(c) Start from a square you know: 25^2 = 625 is too small. Then 26^2 = 676 and 27^2 = 729, so 676 < 700 < 729 and
To find the nearer integer, square the halfway point: 26.5^2 = 702.25, which is more than 700. So \displaystyle \sqrt{700} is below 26.5, in the lower half, and it is nearer to 26.
Closing in on \displaystyle \sqrt{29}
Find \displaystyle \sqrt{29} to the nearest hundredth.
SolutionThe procedure found \displaystyle 5.3 < \sqrt{29} < 5.4. The hundredths in that interval are 5.31 to 5.39, and \displaystyle \sqrt{29} lies between two neighboring ones.
For a first guess, 29 is 0.91 above 5.3^2, out of the 1.07 from 5.3^2 to 5.4^2, so most of the way along. Try 5.38 and 5.39:
The first is below 29 and the second above, so
Now round. The halfway point is 5.385, and 5.385^2 = 28.998225. That is less than 29, only just, so \displaystyle \sqrt{29} is above 5.385, and to the nearest hundredth
Figure 2 shows the two rounds, the second line the shaded tenth of the first, magnified. As a check, 5.39 rounds to 5.4, the nearest tenth found before.
The side of a square garden
A square garden has an area of 90 square meters.
(a) Find the length of its side to the nearest tenth of a meter.
(b) A fence goes all the way round the garden. How much fencing is needed, to the nearest tenth of a meter?
(a) Call the side s meters. The area of a square is its side squared, so s^2 = 90, and since a length is positive, \displaystyle s = \sqrt{90}.
The perfect squares on either side are 81 and 100, so 9 < s < 10. Now 90 is 9 of the 19 steps from 81 to 100, close to halfway, so try 9.5: 9.5^2 = 90.25, just too big. Then 9.4^2 = 88.36 is too small, so
The halfway point is 9.45, and 9.45^2 = 89.3025, which is less than 90. So s is in the upper half, and s \approx 9.5 meters.
(b) The fence is the perimeter, 4s. Multiplying the trap from (a) by 4 gives 37.6 < 4s < 38, which is too wide to give a tenth: the numbers in it round to anything from 37.6 to 38.0. So tighten the trap on s by one more place first:
so 9.48 < s < 9.49. Multiplying by 4 keeps the order:
Every number in this trap rounds to 37.9, so about 37.9 meters of fencing are needed.
Notice that rounding the side first and then multiplying, 4 \times 9.5 = 38.0, gives the wrong tenth. Rounding the side moved it by up to half a tenth, and multiplying by 4 made that up to two tenths. Round once, at the end.
The diagonal of a square tile
A square floor tile has a side of 25 cm. The diagonal of a square is \displaystyle \sqrt{2} times its side, a fact proved in the lesson on Special right triangles, so the diagonal d of this tile is \displaystyle 25\sqrt{2} cm. Find d to the nearest tenth of a centimeter.
First try: use a trap on \displaystyle \sqrt{2}. Since 1.41^2 = 1.9881 and 1.42^2 = 2.0164, we have \displaystyle 1.41 < \sqrt{2} < 1.42. Multiplying all three parts by 25:
That is not enough. The trap on \displaystyle \sqrt{2} was a hundredth long, and multiplying by 25 made it 25 hundredths long, a quarter of a centimeter: numbers in it round to 35.3, 35.4 or 35.5.
Better: square the whole length. By the rule for lengths, d^2 = 25^2 \cdot 2 = 625 \cdot 2 = 1250, so \displaystyle d = \sqrt{1250}. Now trap \displaystyle \sqrt{1250} directly. 35^2 = 1225 and 36^2 = 1296, and then
so 35.3 < d < 35.4. The halfway point gives 35.35^2 = 1249.6225, which is less than 1250, so d is in the upper half:
Every multiplication stretches a trap, so it pays to do the multiplying before taking the square root.
Five irrational numbers on one line
Place \displaystyle \sqrt{3}, \displaystyle \sqrt{7}, \displaystyle 2 + \sqrt{3}, \displaystyle -\sqrt{2} and \displaystyle -\sqrt{14} on one number line, each between two consecutive tenths.
SolutionTrap each square root to tenths, then shift or mirror.
\displaystyle \sqrt{3}: 1^2 = 1 and 2^2 = 4, so it is between 1 and 2. Then 1.7^2 = 2.89 and 1.8^2 = 3.24, so \displaystyle 1.7 < \sqrt{3} < 1.8.
\displaystyle \sqrt{7}: 4 < 7 < 9 puts it between 2 and 3. Then 2.6^2 = 6.76 and 2.7^2 = 7.29, so \displaystyle 2.6 < \sqrt{7} < 2.7.
\displaystyle 2 + \sqrt{3}: add 2 to every part of the trap on \displaystyle \sqrt{3}. Adding keeps the order, so \displaystyle 3.7 < 2 + \sqrt{3} < 3.8. On the line it is \displaystyle \sqrt{3} moved 2 to the right.
\displaystyle -\sqrt{2}: 1.4^2 = 1.96 and 1.5^2 = 2.25, so \displaystyle 1.4 < \sqrt{2} < 1.5. Multiplying by -1 reverses the order and swaps the ends: \displaystyle -1.5 < -\sqrt{2} < -1.4.
\displaystyle -\sqrt{14}: 9 < 14 < 16 puts \displaystyle \sqrt{14} between 3 and 4. Then 3.7^2 = 13.69 and 3.8^2 = 14.44, so \displaystyle 3.7 < \sqrt{14} < 3.8, and mirroring gives \displaystyle -3.8 < -\sqrt{14} < -3.7.
Read from the line, the order from least to greatest is \displaystyle -\sqrt{14}, \displaystyle -\sqrt{2}, \displaystyle \sqrt{3}, \displaystyle \sqrt{7}, \displaystyle 2 + \sqrt{3}.
Putting five numbers in order
Put these numbers in order from least to greatest:
Get each one between two numbers you can compare.
\pi = 3.14159\ldots, so \pi < 3.15. Dividing 19 by 6 gives \displaystyle \frac{19}{6} = 3.1666\ldots, which is above 3.15 and below 3.2. So \displaystyle \pi < \frac{19}{6}.
\displaystyle 1 + \sqrt{5}: 2.2^2 = 4.84 and 2.3^2 = 5.29, so \displaystyle 2.2 < \sqrt{5} < 2.3, and adding 1 gives \displaystyle 3.2 < 1 + \sqrt{5} < 3.3. That puts it above \displaystyle \frac{19}{6} and below 3.3.
\displaystyle \sqrt{11} against 3.3: both are positive, so compare squares. 3.3^2 = 10.89, which is less than 11, so \displaystyle 3.3 < \sqrt{11}.
Putting the pieces together:
Three of the five are irrational and two are not; the method did not need to know which.
Estimating \pi^2, and a close contest
(a) Using 3.14 < \pi < 3.15, find \pi^2 to the nearest tenth.
(b) Which is larger, \displaystyle \sqrt{2} + \sqrt{3} or \pi?
Solution(a) All three numbers in 3.14 < \pi < 3.15 are positive, so squaring keeps the order:
Every number in this trap rounds to 9.9, so \pi^2 \approx 9.9.
(b) Start with traps to hundredths. \displaystyle 1.41 < \sqrt{2} < 1.42, and since 1.73^2 = 2.9929 and 1.74^2 = 3.0276, \displaystyle 1.73 < \sqrt{3} < 1.74. Adding the ends,
That does not decide it, because \pi = 3.14159\ldots is inside this trap too. Tighten to thousandths. 1.414^2 = 1.999396 and 1.415^2 = 2.002225, so \displaystyle \sqrt{2} > 1.414. And 1.732^2 = 2.999824, which is less than 3, so \displaystyle \sqrt{3} > 1.732. Adding,
and \pi < 3.142. The traps no longer overlap, so \displaystyle \sqrt{2} + \sqrt{3} > \pi.
The two differ by less than a hundredth, which is why the first round could not tell them apart. You narrow until the traps separate, and two different numbers always separate eventually.
Boundaries
The squaring test needs positive numbers. -3 < 2, but (-3)^2 = 9 is larger than 2^2 = 4. With negatives, compare the positive versions first and then reverse, as in Example 5.
A real square root needs a number that is not negative. \displaystyle \sqrt{-4} is not on the number line at all, since no real number squares to -4. And \displaystyle \sqrt{n} always means the root that is not negative; the other one is \displaystyle -\sqrt{n}.
Between 0 and 1 the integer trap tells you little. Every number between 0 and 1 has its square root between 0 and 1 too, so trap it with squares of tenths instead: 0.25 < 0.3 < 0.36 gives \displaystyle 0.5 < \sqrt{0.3} < 0.6. Notice that \displaystyle \sqrt{0.3} is larger than 0.3: for numbers between 0 and 1, taking the square root makes them bigger.
The decimal is an approximation, never the number. \displaystyle \sqrt{29} \approx 5.39, but \displaystyle \sqrt{29} = 5.39 is false: 5.39^2 = 29.0521.
A trap can straddle a rounding point. In Example 4, 35.25 < d < 35.5 held numbers that round to three different tenths, so the estimate was not yet known. The fix is one more decimal place, not a guess. Combining traps makes this more likely: multiplying a trap by 25 makes it 25 times as long, and adding two traps adds their lengths.
Products of traps need positive numbers. ac < xy < bd holds when a and c are positive. If a factor can be negative, the ends can swap or the trap can even run through 0.
Where this goes next
Trapping an irrational number between decimals as close together as you like is what lets it take its place on the real number line, side by side with the rationals, and it feeds deciding which set a number belongs to.
Lengths like the garden’s side and the tile’s diagonal come up constantly with the Pythagorean theorem, which hands you the square of a length and asks for the length; simplifying the square root links \displaystyle \sqrt{1250} and \displaystyle 25\sqrt{2}. The same trapping works with cubes for Cube roots and higher roots, and Newton’s method closes in faster, roughly doubling the correct digits each round.
Exercises
Part 1: between two integers
Between which two consecutive integers does each number lie?
- \displaystyle \sqrt{43}
- \displaystyle \sqrt{110}
- \displaystyle \sqrt{250}
- \displaystyle \sqrt{2000}
Part 1: closer and closer
Find each square root to the place asked for, by squaring trial values.
- \displaystyle \sqrt{13}, to the nearest hundredth
- \displaystyle \sqrt{61}, to the nearest tenth
- \displaystyle \sqrt{21}, to the nearest hundredth
- \displaystyle \sqrt{320}, to the nearest tenth
Part 1: lengths
- A square garden has an area of 45 square meters. Find its side to the nearest tenth of a meter.
- A square tile has a side of 15 cm, so its diagonal is \displaystyle 15\sqrt{2} cm. Find the diagonal to the nearest tenth of a centimeter.
- A rectangular doormat is twice as long as it is wide, and its area is 3000 square centimeters. Find its width to the nearest tenth of a centimeter.
Part 2: placing on the line
- Give each number between two consecutive tenths: \displaystyle \sqrt{33}, \displaystyle -\sqrt{33} and \displaystyle 6 - \sqrt{33}.
- Between which two consecutive integers does each number lie? (a) \displaystyle 4 + \sqrt{40} (b) \displaystyle 10 - \sqrt{40} (c) \displaystyle 2\sqrt{40}
Part 2: comparing and ordering
- Which is larger, \displaystyle \sqrt{22} or 4.7?
- Which is larger, \displaystyle 3\sqrt{3} or 5?
- Put in order from least to greatest: \displaystyle \sqrt{19}, 4.36, 4.35, \displaystyle \frac{48}{11}.
- Put in order from least to greatest: \displaystyle -\sqrt{15}, -4, -3.8, -\pi.
Part 2: estimating expressions
- Find \displaystyle \sqrt{3} + \sqrt{12} to the nearest tenth, starting from traps to hundredths.
- Using 3.141 < \pi < 3.142 and \displaystyle 1.414 < \sqrt{2} < 1.415, find \displaystyle \pi\sqrt{2} to the nearest tenth.
- Which is larger, \displaystyle \sqrt{2} + \sqrt{5} or \displaystyle \sqrt{13}?
Reasoning
- Explain why, if n is a whole number strictly between the perfect squares k^2 and (k + 1)^2, the number \displaystyle \sqrt{n} cannot be a whole number.
- Prove that for positive numbers a and b, a < b exactly when a^2 < b^2. Then explain why this lets you test a trial value for a square root by squaring it.
- Augustin says that since 50 lies between 49 and 64, \displaystyle \sqrt{50} is about 7.5. Explain what is wrong, and find \displaystyle \sqrt{50} to the nearest tenth.
- Given \displaystyle 4.7 < \sqrt{23} < 4.8, Bisi writes \displaystyle 8 - 4.7 < 8 - \sqrt{23} < 8 - 4.8. Explain what is wrong, and give the correct trap for \displaystyle 8 - \sqrt{23}.
- Explain why no decimal that stops, such as 5.385 or 5.3851647, can square to exactly 29.
Answers
- \displaystyle 6 < \sqrt{43} < 7, since 36 < 43 < 49.
- \displaystyle 10 < \sqrt{110} < 11, since 100 < 110 < 121.
- \displaystyle 15 < \sqrt{250} < 16, since 225 < 250 < 256.
- \displaystyle 44 < \sqrt{2000} < 45, since 44^2 = 1936 and 45^2 = 2025.
- \displaystyle \sqrt{13} \approx 3.61. 3.6^2 = 12.96 and 3.61^2 = 13.0321, so \displaystyle 3.60 < \sqrt{13} < 3.61; the halfway point gives 3.605^2 = 12.996025 < 13, so round up.
- \displaystyle \sqrt{61} \approx 7.8. 7.8^2 = 60.84 and 7.9^2 = 62.41; the halfway point gives 7.85^2 = 61.6225 > 61, so round down.
- \displaystyle \sqrt{21} \approx 4.58. 4.58^2 = 20.9764 and 4.59^2 = 21.0681; the halfway point gives 4.585^2 = 21.022225 > 21, so round down.
- \displaystyle \sqrt{320} \approx 17.9. 17.8^2 = 316.84 and 17.9^2 = 320.41; the halfway point gives 17.85^2 = 318.6225 < 320, so round up.
- About 6.7 meters. The side is \displaystyle \sqrt{45}; 6.7^2 = 44.89 and 6.8^2 = 46.24, and 6.75^2 = 45.5625 > 45.
- About 21.2 cm. The diagonal is \displaystyle \sqrt{15^2 \cdot 2} = \sqrt{450}; 21.2^2 = 449.44 and 21.3^2 = 453.69, and 21.25^2 = 451.5625 > 450.
- About 38.7 cm. If the width is w, the length is 2w and 2w \cdot w = 3000, so w^2 = 1500. Then 38.7^2 = 1497.69 and 38.8^2 = 1505.44, and 38.75^2 = 1501.5625 > 1500.
- \displaystyle 5.7 < \sqrt{33} < 5.8, since 5.7^2 = 32.49 and 5.8^2 = 33.64. Mirroring, \displaystyle -5.8 < -\sqrt{33} < -5.7, and adding 6 to that, \displaystyle 0.2 < 6 - \sqrt{33} < 0.3.
- From \displaystyle 6 < \sqrt{40} < 7, since 36 < 40 < 49: (a) \displaystyle 10 < 4 + \sqrt{40} < 11. (b) \displaystyle 3 < 10 - \sqrt{40} < 4, the ends crossing over. (c) \displaystyle 12 < 2\sqrt{40} < 13: doubling the trap gives only 12 to 14, so square instead, \displaystyle 2\sqrt{40} = \sqrt{160} and 144 < 160 < 169.
- 4.7 is larger: 4.7^2 = 22.09 > 22.
- \displaystyle 3\sqrt{3} is larger: \displaystyle (3\sqrt{3})^2 = 9 \cdot 3 = 27 > 25.
- \displaystyle 4.35 < \sqrt{19} < 4.36 < \frac{48}{11}. 4.35^2 = 18.9225 < 19 and 4.36^2 = 19.0096 > 19, and \displaystyle \frac{48}{11} = 4.3636\ldots
- \displaystyle -4 < -\sqrt{15} < -3.8 < -\pi. For the positive versions, 3.8^2 = 14.44 < 15 < 16 gives \displaystyle 3.8 < \sqrt{15} < 4, and \pi < 3.8; so \displaystyle \pi < 3.8 < \sqrt{15} < 4, and negating reverses the whole order.
- \displaystyle \sqrt{3} + \sqrt{12} \approx 5.2. \displaystyle 1.73 < \sqrt{3} < 1.74, and 3.46^2 = 11.9716 and 3.47^2 = 12.0409 give \displaystyle 3.46 < \sqrt{12} < 3.47. Adding, \displaystyle 5.19 < \sqrt{3} + \sqrt{12} < 5.21, and all of it rounds to 5.2.
- \displaystyle \pi\sqrt{2} \approx 4.4. All the numbers are positive, so multiply the ends: 3.141 \times 1.414 = 4.441374 and 3.142 \times 1.415 = 4.44593, giving \displaystyle 4.441374 < \pi\sqrt{2} < 4.44593, all of which rounds to 4.4.
- \displaystyle \sqrt{2} + \sqrt{5} is larger. \displaystyle \sqrt{2} > 1.41 and, since 2.23^2 = 4.9729 < 5, \displaystyle \sqrt{5} > 2.23, so \displaystyle \sqrt{2} + \sqrt{5} > 3.64. But 3.61^2 = 13.0321 > 13, so \displaystyle \sqrt{13} < 3.61.
- From k^2 < n < (k + 1)^2 and the squaring fact, \displaystyle k < \sqrt{n} < k + 1. There is no whole number strictly between two consecutive whole numbers, so \displaystyle \sqrt{n} is not one.
- If 0 < a < b, multiplying by a and by b gives a^2 < ab and ab < b^2, so a^2 < b^2. If instead b \le a, the same steps give b^2 \le a^2, so a^2 < b^2 forces a < b. A trial value t > 0 is below \displaystyle \sqrt{n} exactly when t^2 < n, since \displaystyle (\sqrt{n})^2 = n; so squaring t tells you which side of \displaystyle \sqrt{n} it is on.
- 7.5^2 = 56.25, far above 50, so \displaystyle \sqrt{50} is well below 7.5. Augustin took the middle of 7 and 8, but 50 is only 1 above 49 and 14 below 64, so it is near the bottom of the interval. 7.1^2 = 50.41 > 50, so \displaystyle 7 < \sqrt{50} < 7.1, and 7.05^2 = 49.7025 < 50 gives \displaystyle \sqrt{50} \approx 7.1.
- Subtracting from 8 reverses the order, so the ends must swap: \displaystyle 8 - 4.8 < 8 - \sqrt{23} < 8 - 4.7, that is \displaystyle 3.2 < 8 - \sqrt{23} < 3.3. Bisi’s version says \displaystyle 3.3 < 8 - \sqrt{23} < 3.2, which no number satisfies.
- A decimal that stops is a fraction, such as \displaystyle 5.385 = \frac{5385}{1000}. If one squared to exactly 29, it would be \displaystyle \sqrt{29}, making \displaystyle \sqrt{29} a fraction. But 29 is a whole number that is not a perfect square, so \displaystyle \sqrt{29} is irrational and is not a fraction.