---
title: "Square Roots"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 65
exercises: 10
source: https://one-course.com/books/math/1/en/chapter/65-square-roots
---

# Chapter 65 — Square Roots

How long is the diagonal of a square of side $1$? The Pythagorean theorem answers $\sqrt 2$, a number whose square is $2$ — and which turns out not to be a [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction). This chapter defines [square roots](#def-g9-sqrt-def), establishes the rules for computing with them, and teaches how to simplify expressions like $\sqrt{75}$.

## 65.1 Definition and first properties

**Definition 65.1 (Square root).**

Let $a \geq 0$. The *square root* of $a$, written $\sqrt a$, is the unique *nonnegative* number whose square is $a$:

$$
\sqrt a \geq 0
\qquad\text{and}\qquad
\left(\sqrt a\right)^2 = a .
$$

Negative numbers have no square root, since every square is nonnegative.

**Example 65.2.**

$\sqrt{49} = 7$, $\sqrt 0 = 0$, $\sqrt 1 = 1$, $\sqrt{2.25} = 1.5$. The first [square roots](#def-g9-sqrt-def) to know by heart are those of the *perfect squares*: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121,
144$.

![The diagonal of a unit square has length √12 + 12 = √2 1.414, by the Pythagorean theorem ( recalls it). This number is irrational: its decimals never repeat.](https://one-course.com/images/onecourse/chapters/math-1/g9-sqrt/fig-e4c870763528.svg)

*The diagonal of a unit square has length $\sqrt{1^2 + 1^2} =
\sqrt2 \approx 1.414$, by the Pythagorean theorem ([Chapter 69](https://one-course.com/books/math/1/en/chapter/69-trigonometry-in-the-right-triangle#ch-g9-trig) recalls it). This number is irrational: its decimals never repeat.*

**Proposition 65.3 (Square root of a square).**

For every real number $x$ (positive or not):

$$
\sqrt{x^2} = \abs{x} =
\begin{cases}
x & \text{if } x \geq 0,\\
-x & \text{if } x < 0 .
\end{cases}
$$

**Proof.** The number $\abs{x}$ is nonnegative and its square is $x^2$ (a number and its opposite have the same square). Being the unique nonnegative number of square $x^2$, it is $\sqrt{x^2}$. ∎

**Example 65.4.**

$\sqrt{(-5)^2} = \sqrt{25} = 5 = \abs{-5}$: the [square root](#def-g9-sqrt-def) “forgets” the sign, it does not restore it. In particular $\sqrt{x^2} = x$ is *wrong* for negative $x$.

## 65.2 Products and quotients of square roots

**Theorem 65.5 (Multiplication and division rules).**

For all $a \geq 0$ and $b \geq 0$:

$$
\sqrt{a b} = \sqrt a \times \sqrt b,
\qquad\text{and for } b > 0:\quad
\sqrt{\frac ab} = \frac{\sqrt a}{\sqrt b} .
$$

**Proof.** The number $\sqrt a \times \sqrt b$ is nonnegative ([product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of nonnegative numbers), and its square is

$$
\left(\sqrt a \times \sqrt b\right)^2
= \left(\sqrt a\right)^2 \times \left(\sqrt b\right)^2
= a b .
$$

By uniqueness of the nonnegative number whose square is $ab$, it equals $\sqrt{ab}$. The [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) rule is proved the same way. ∎

**Remark 65.6 (No such rule for sums!).**

$\sqrt{a + b}$ is *not* $\sqrt a + \sqrt b$ in general:

$$
\sqrt{9 + 16} = \sqrt{25} = 5,
\qquad\text{but}\qquad
\sqrt 9 + \sqrt{16} = 3 + 4 = 7 .
$$

**Method 65.7 (Simplifying n\sqrt nn​).**

To simplify the [square root](#def-g9-sqrt-def) of an integer:

1. find the largest perfect square dividing $n$ (factor $n$ if needed);
2. split with the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) rule and extract the square: $\sqrt{k^2 m} = k\sqrt m$ ;
3. check that no perfect square [divides](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) what is left under the root.

**Example 65.8.**

Simplify $\sqrt{75}$: since $75 = 25 \times 3$,

$$
\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt 3 = 5\sqrt3 .
$$

Simplify $\sqrt{72}$: since $72 = 36 \times 2$, $\sqrt{72} = 6\sqrt2$. And a [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder): $\sqrt{\dfrac{49}{16}} = \dfrac{\sqrt{49}}{\sqrt{16}} = \dfrac74$.

**Example 65.9 (Adding square roots).**

[Sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of [square roots](#def-g9-sqrt-def) simplify only when the roots are *alike*. Compute $\sqrt{12} + \sqrt{27} - \sqrt{48}$, simplifying each term first:

$$
\begin{align*}
\sqrt{12} &= \sqrt{4 \times 3} = 2\sqrt3, \\
\sqrt{27} &= \sqrt{9 \times 3} = 3\sqrt3, \\
\sqrt{48} &= \sqrt{16 \times 3} = 4\sqrt3,
\end{align*}
$$

so the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is $2\sqrt3 + 3\sqrt3 - 4\sqrt3 = (2 + 3 - 4)\sqrt3 = \sqrt3$.

**Example 65.10 (Expanding with square roots).**

The identities of algebra apply to [square roots](#def-g9-sqrt-def). Expand $\left(\sqrt5 + 2\right)^2$ with $(a+b)^2 = a^2 + 2ab + b^2$:

$$
\left(\sqrt5 + 2\right)^2 = 5 + 2 \times 2\sqrt5 + 4 = 9 + 4\sqrt5 .
$$

And with the third identity, $(a+b)(a-b) = a^2 - b^2$:

$$
\left(\sqrt7 + \sqrt3\right)\left(\sqrt7 - \sqrt3\right) = 7 - 3 = 4 :
$$

the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of two irrational numbers can be an integer.

## 65.3 The equation $x^2 = a$

**Theorem 65.11 (Solving x2=ax^2 = ax2=a).**

1. If $a > 0$ , the [equation](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) $x^2 = a$ has exactly two solutions: $\sqrt a$ and $-\sqrt a$ ;
2. if $a = 0$ , the only solution is $0$ ;
3. if $a < 0$ , there is no solution.

**Proof.** Rewrite $x^2 = a$ as $x^2 - \left(\sqrt a\right)^2 = 0$ when $a \geq 0$, and factor as a [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of squares:

$$
\left(x - \sqrt a\right)\left(x + \sqrt a\right) = 0 ,
$$

so $x = \sqrt a$ or $x = -\sqrt a$ (these coincide when $a = 0$). When $a < 0$ there is no solution, since $x^2 \geq 0 > a$ for every $x$. ∎

![Solving x2 = 3 on the parabola y = x2: the horizontal line y = 3 cuts it at the two abscissas ±√3.](https://one-course.com/images/onecourse/chapters/math-1/g9-sqrt/fig-57ab6201daa2.svg)

*Solving $x^2 = 3$ on the parabola $y = x^2$: the horizontal line $y = 3$ cuts it at the two abscissas $\pm\sqrt3$.*

**Example 65.12.**

$x^2 = 16$: solutions $4$ and $-4$. $x^2 = 5$: solutions $\sqrt5$ and $-\sqrt5$ (exact values; $\sqrt5
\approx 2.236$). $x^2 = -9$: no solution. $3x^2 = 21$: divide by 3 first, $x^2 = 7$, solutions $\pm\sqrt7$.

## 65.4 Exercises

**Exercise 65.1 ★.**

Compute without a calculator:

$$
\sqrt{64}, \qquad \sqrt{100}, \qquad \sqrt{0.09}, \qquad
\sqrt{\tfrac{1}{25}}, \qquad \left(\sqrt{13}\right)^2, \qquad
\sqrt{(-4)^2} .
$$

**Solution of Exercise 65.1.**

$\sqrt{64} = 8$; $\sqrt{100} = 10$; $\sqrt{0.09} = 0.3$ (since $0.3^2 = 0.09$); $\sqrt{\frac{1}{25}} = \frac15$; $\left(\sqrt{13}\right)^2 = 13$; $\sqrt{(-4)^2} = \sqrt{16} = 4 = \abs{-4}$.

**Exercise 65.2 ★.**

Simplify:

$$
\sqrt{50}, \qquad \sqrt{45}, \qquad \sqrt{98}, \qquad \sqrt{300} .
$$

**Solution of Exercise 65.2.**

$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2$; $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt5$; $\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt2$; $\sqrt{300} = \sqrt{100 \times 3} = 10\sqrt3$.

**Exercise 65.3 ★.**

Compute and simplify:

$$
\sqrt2 \times \sqrt{18}, \qquad
\frac{\sqrt{75}}{\sqrt3}, \qquad
\sqrt5 \times \sqrt{20}, \qquad
\frac{\sqrt{8}}{\sqrt{50}} .
$$

**Solution of Exercise 65.3.**

$\sqrt2 \times \sqrt{18} = \sqrt{36} = 6$.

$\dfrac{\sqrt{75}}{\sqrt3} = \sqrt{\dfrac{75}{3}} = \sqrt{25} = 5$.

$\sqrt5 \times \sqrt{20} = \sqrt{100} = 10$.

$\dfrac{\sqrt8}{\sqrt{50}} = \sqrt{\dfrac{8}{50}} = \sqrt{\dfrac{4}{25}}
= \dfrac25$.

**Exercise 65.4 ★.**

Solve: $x^2 = 36$; $x^2 = 11$; $x^2 + 4 = 0$; $2x^2 = 50$.

**Solution of Exercise 65.4.**

$x^2 = 36$: $x = 6$ or $x = -6$.

$x^2 = 11$: $x = \sqrt{11}$ or $x = -\sqrt{11}$.

$x^2 + 4 = 0$ means $x^2 = -4 < 0$: no solution.

$2x^2 = 50$: $x^2 = 25$, so $x = 5$ or $x = -5$.

**Exercise 65.5 ★.**

Reduce to a single term: $\sqrt{20} + \sqrt{45}$, then $3\sqrt8 - \sqrt{18} + \sqrt2$.

**Solution of Exercise 65.5.**

$\sqrt{20} + \sqrt{45} = 2\sqrt5 + 3\sqrt5 = 5\sqrt5$.

$3\sqrt8 - \sqrt{18} + \sqrt2 = 3 \times 2\sqrt2 - 3\sqrt2 + \sqrt2
= (6 - 3 + 1)\sqrt2 = 4\sqrt2$.

**Exercise 65.6 ★★.**

Expand and simplify:

$$
\left(\sqrt3 + 1\right)^2, \qquad
\left(2\sqrt5 - 3\right)^2, \qquad
\left(\sqrt6 + \sqrt2\right)\left(\sqrt6 - \sqrt2\right).
$$

**Solution of Exercise 65.6.**

$\left(\sqrt3 + 1\right)^2 = 3 + 2\sqrt3 + 1 = 4 + 2\sqrt3$.

$\left(2\sqrt5 - 3\right)^2 = 4 \times 5 - 2 \times 3 \times 2\sqrt5 + 9
= 29 - 12\sqrt5$.

$\left(\sqrt6 + \sqrt2\right)\left(\sqrt6 - \sqrt2\right) = 6 - 2 = 4$.

**Exercise 65.7 ★★.**

A square field has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $6400$ m$^2$. What is the length of its side? Of its diagonal (exact value, then rounded to the nearest meter)?

**Solution of Exercise 65.7.**

Side: $\sqrt{6400} = 80$ m. Diagonal (Pythagoras): $\sqrt{80^2 + 80^2} = 80\sqrt2 \approx 113$ m.

**Exercise 65.8 ★★.**

Show that $\dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}$ (multiply [numerator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) and [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) by $\sqrt2$). Use the same trick to write $\dfrac{6}{\sqrt3}$ and $\dfrac{10}{\sqrt5}$ without a [square root](#def-g9-sqrt-def) in the [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def).

**Solution of Exercise 65.8.**

$\dfrac{1}{\sqrt2} = \dfrac{1 \times \sqrt2}{\sqrt2 \times \sqrt2}
= \dfrac{\sqrt2}{2}$.

$\dfrac{6}{\sqrt3} = \dfrac{6\sqrt3}{3} = 2\sqrt3$. $\dfrac{10}{\sqrt5} = \dfrac{10\sqrt5}{5} = 2\sqrt5$.

**Exercise 65.9 ★★.**

True or false? Justify with a proof or a counterexample.

1. For all $a, b \geq 0$ : $\sqrt{ab} = \sqrt a \sqrt b$ .
2. For all $a, b \geq 0$ : $\sqrt{a+b} = \sqrt a + \sqrt b$ .
3. For all $x$ : $\sqrt{x^2} = x$ .
4. $\left(3\sqrt2\right)^2 = 18$ .

**Solution of Exercise 65.9.**

*1. True*: this is the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) rule ([Theorem 65.5](#thm-g9-sqrt-rules)).

*2. False*: $\sqrt{9 + 16} = 5$ but $\sqrt9 + \sqrt{16} = 7$.

*3. False* for negative $x$: $\sqrt{(-4)^2} = 4 \neq -4$. The correct statement is $\sqrt{x^2} = \abs{x}$.

*4. True*: $\left(3\sqrt2\right)^2 = 9 \times 2 = 18$.

**Exercise 65.10 ★★★.**

Let $x = \sqrt{7 + 4\sqrt3}$. Compute $\left(2 + \sqrt3\right)^2$, and deduce a simpler expression of $x$. Same question for $\sqrt{9 - 4\sqrt5}$ (aim for a square of the form $\left(\sqrt5 - 2\right)^2$, and mind the sign).

**Solution of Exercise 65.10.**

$\left(2 + \sqrt3\right)^2 = 4 + 4\sqrt3 + 3 = 7 + 4\sqrt3$. So $x = \sqrt{7 + 4\sqrt3} = \sqrt{\left(2+\sqrt3\right)^2} = 2 + \sqrt3$ (a nonnegative number, so the root just removes the square).

Similarly $\left(\sqrt5 - 2\right)^2 = 5 - 4\sqrt5 + 4 = 9 - 4\sqrt5$, and $\sqrt5 - 2 > 0$, so $\sqrt{9 - 4\sqrt5} = \sqrt5 - 2$ (not $2 - \sqrt5$, which is negative!).

## 65.5 Problem: The number that is not a fraction

**Problem 65.1.**

Weekend problem — the legendary proof that $\sqrt2$ is irrational, its many cousins, and Heron’s uncannily good recipe for approximating it

The diagonal of a unit square is a perfectly real length — you drew it in [Example 58.9](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#ex-g8-pythagoras-diagonal) — and yet, as the Pythagoreans discovered to their horror some twenty-five centuries ago, *no [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) whatsoever* measures it. Legend claims the discovery was punished by drowning. This problem walks you through the immortal proof (four questions and it is yours for life), multiplies the victims, and ends with the recipe engineers used for two thousand years to tame the untameable number.

**Part I — The proof.**

1. First hunt it down: check that $1.4 < \sqrt2 < 1.5$ , then that $1.41 < \sqrt2 < 1.42$ , by squaring the bounds. One more digit: between which three-decimal numbers does $\sqrt2$ lie?
2. Now suppose — for the sake of contradiction — that $\sqrt2 = \frac ab$ for some [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) *in lowest terms* ( [Method 64.16](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#met-g9-arith-simplify) ). Square both sides and show that $a^2 = 2 b^2$ . What is the parity of $a^2$ ?
3. The parity facts of [Problem 58.1](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#pb-g8-pythagoras-1) say: [odd numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) have [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) squares. Deduce that $a$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) , write $a = 2k$ , substitute — and show that $b$ must be [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) too.
4. Where is the contradiction? Conclude, and state the theorem in full: *$\sqrt2$ is irrational — it is no [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) of whole numbers* .
5. The proof leaned once, discreetly, on the words “in lowest terms”. Point to the exact step that would collapse without them, and explain why assuming lowest terms was legitimate in the first place.

**Part II — The victims multiply.**

6. A second proof style, via prime factorizations ( [Theorem 64.6](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#thm-g9-arith-factorization) ): in $a^2 = 3b^2$ , compare the parity of the *[exponent](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def) of $3$* on each side ( [Problem 64.1](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#pb-g9-arith-1) : squares carry [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) [exponents](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def) ). Conclude that $\sqrt3$ is irrational.
7. Run the same [exponent](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def) argument on $a^2 = n b^2$ for a general whole number $n$ : for which $n$ does it produce a contradiction, and for which does it fail? State the complete result: $\sqrt n$ is irrational exactly when …
8. Prove the shield lemma: a nonzero rational times an irrational is irrational. (Suppose $r x = q$ with $r, q$ rational, $r \neq 0$ , and solve for $x$ .) Deduce that $2\sqrt2$ and $\frac{\sqrt2}{2}$ are irrational.
9. Prove that $1 + \sqrt2$ is irrational. Then the pretty one: supposing $s = \sqrt2 + \sqrt3$ were rational, compute $s^2$ , isolate $\sqrt6$ , and find the contradiction.
10. Temper the enthusiasm: give two irrational numbers whose *[sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def)* is rational, and two whose *[product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def)* is rational. (Irrationality is not preserved by arithmetic — each case needs its own proof.)

**Part III — Heron’s recipe.** Two thousand years before calculators, Heron of Alexandria approximated $\sqrt2$ like this: *guess $x$; replace the guess by the average of $x$ and $\frac2x$; repeat.*

11. Start from the guess $x = 1$ and compute the next two guesses as exact [fractions](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) .
12. Compute the third guess, again as an exact [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) , and its decimal value. Compare with question 1: how many decimals of $\sqrt2$ are already correct?
13. The idea behind the recipe: a rectangle of [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $2$ with one side $x$ has the other side $\frac2x$ . Show that $\sqrt2$ always lies *between* $x$ and $\frac2x$ (consider the two cases $x^2 > 2$ and $x^2 < 2$ ), so averaging the two sides squeezes the rectangle towards the square.
14. The guesses $\frac32$ , $\frac{17}{12}$ , $\frac{577}{408}$ hide a gem: compute $3^2 - 2 \times  2^2$ , then $17^2 - 2 \times 12^2$ , then $577^2 - 2 \times 408^2$ . Part I proved $a^2 - 2b^2 = 0$ impossible — how close do Heron’s [fractions](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) come to the impossible?
15. Finale, in two sentences: the diagonal of the unit square exists on paper, no [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) measures it, and its decimal writing can never repeat ( [Problem 63.1](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#pb-g9-fractions-1) ). What kind of “new numbers” must the number line therefore contain — and where in this series is their full story told?

**Solution of Problem 65.1.**

**1.** $1.4^2 = 1.96 < 2 < 2.25 = 1.5^2$; $1.41^2 = 1.9881 < 2 < 2.0164 = 1.42^2$. Three decimals: $1.414^2 = 1.999396 < 2 < 2.002225 = 1.415^2$, so $1.414 < \sqrt2 < 1.415$.

**2.** Squaring $\sqrt2 = \frac ab$ gives $2 = \frac{a^2}{b^2}$, hence $a^2 = 2b^2$: the number $a^2$ is twice a whole number — *[even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)*.

**3.** If $a$ were [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), $a^2$ would be [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) ([Problem 58.1](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#pb-g8-pythagoras-1)); since $a^2$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), $a$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd): $a = 2k$. Substituting: $4k^2 = 2b^2$, so $b^2 = 2k^2$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), and by the same parity fact $b$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd).

**4.** Both $a$ and $b$ [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) means both [divisible](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $2$ — but $\frac ab$ was in lowest terms, sharing no common [divisor](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor). Contradiction: the assumed [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) cannot exist. *Theorem: $\sqrt2$ is irrational.*

**5.** The contradiction lives entirely in “lowest terms”: without it, “$a$ and $b$ both [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)” contradicts nothing. Assuming lowest terms is legitimate because every [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) *has* a lowest-terms form ([Method 64.16](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#met-g9-arith-simplify): divide out the GCD) — if $\sqrt2$ were any [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) at all, it would be a lowest-terms one, and that one is impossible.

**6.** In prime factorizations, squares carry [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) [exponents](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def) ([Problem 64.1](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#pb-g9-arith-1)). In $a^2 = 3b^2$, the [exponent](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def) of $3$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) on the left, but [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) on the right ([even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) from $b^2$, plus one). No number has two factorizations ([Theorem 64.6](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#thm-g9-arith-factorization)): contradiction, and $\sqrt3$ is irrational.

**7.** In $a^2 = n b^2$, the argument finds a prime with contradictory [exponent](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def) parity exactly when some prime appears in $n$ with an *[odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)* [exponent](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def). If every [exponent](https://one-course.com/books/math/1/en/chapter/56-powers#def-g8-powers-def) of $n$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), $n$ is a perfect square ($n = m^2$, and $\sqrt n = m$ is whole). Complete result: $\sqrt n$ is irrational for every whole $n$ that is not a perfect square — $\sqrt2, \sqrt3, \sqrt5, \sqrt6, \sqrt7, \sqrt8, \sqrt{10},
\dots$

**8.** From $r x = q$ with $r \neq 0$: $x = \frac qr$, a [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) of rationals, hence rational. So if $x$ is irrational, $r x$ cannot be rational: $2\sqrt2$ and $\frac{\sqrt2}{2}$ are irrational (multipliers $2$ and $\frac12$).

**9.** If $1 + \sqrt2 = q$ were rational, then $\sqrt2 = q - 1$ would be rational: contradiction. If $s = \sqrt2 + \sqrt3$ were rational, then $s^2 = 2 + 2\sqrt6 + 3 = 5 + 2\sqrt6$ would be too, giving $\sqrt6 = \frac{s^2 - 5}{2}$ rational — but $\sqrt6$ is irrational (question 7). So $\sqrt2 + \sqrt3$ is irrational.

**10.** [Sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def): $\sqrt2$ and $-\sqrt2$ (or $\sqrt2$ and $1 - \sqrt2$) are each irrational, with rational [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $0$ and $1$. [Products](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def): $\sqrt2 \times \sqrt2 = 2$. Irrationality can evaporate in arithmetic — hence the case-by-case proofs.

**11.** From $x = 1$: average of $1$ and $\frac21 = 2$: $x_1 = \frac32$. Then average of $\frac32$ and $\frac{2}{3/2} = \frac43$: $x_2 = \frac12\left(\frac32 + \frac43\right)
= \frac12 \cdot \frac{17}{6} = \frac{17}{12} \approx 1.4167$.

**12.** $x_3 = \frac12\left(\frac{17}{12} +
\frac{24}{17}\right) = \frac12 \cdot
\frac{289 + 288}{204} = \frac{577}{408} = 1.4142156\ldots$ Against $\sqrt2 = 1.4142135\ldots$: five decimals correct after three steps — the recipe roughly doubles the correct decimals each [round](https://one-course.com/books/math/1/en/chapter/21-large-numbers#def-g4-numbers-round).

**13.** If $x^2 > 2$ (guess too big), then $x > \sqrt2$, and $\frac2x < \frac{2}{\sqrt2} = \sqrt2$: the companion side is too small. If $x^2 < 2$, the inequalities reverse. Either way $\sqrt2$ sits between $x$ and $\frac2x$, so their average — the next guess — is closer than the worse of the two sides: the rectangle of [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $2$ gets squarer at every step.

**14.** $9 - 8 = 1$; $289 - 288 = 1$; $332\,929 - 332\,928 = 1$. Every Heron guess $\frac ab$ satisfies $a^2 - 2b^2 = 1$: Part I proved that hitting $0$ is impossible, and these [fractions](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) miss the impossible by exactly one unit — the closest whole numbers can ever come.

**15.** The number line must contain numbers beyond the [fractions](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) — lengths like $\sqrt2$, whose decimal writings run forever without repeating: the *real numbers*. Their honest construction is a story for the High School volume and, in full rigor, the university ones.