65Square Roots

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.

3. If $a$ were odd, $a^2$ would be odd (Problem 58.1); since $a^2$ is even, $a$ is even: $a = 2k$. Substituting: $4k^2 = 2b^2$, so $b^2 = 2k^2$ is even, and by the same parity fact $b$ is even.

4. Both $a$ and $b$ even means both divisible by $2$ — but $\frac ab$ was in lowest terms, sharing no common divisor. Contradiction: the assumed fraction cannot exist. Theorem: $\sqrt2$ is irrational.

5. The contradiction lives entirely in “lowest terms”: without it, “$a$ and $b$ both even” contradicts nothing. Assuming lowest terms is legitimate because every fraction has a lowest-terms form (Method 64.16: divide out the GCD) — if $\sqrt2$ were any fraction at all, it would be a lowest-terms one, and that one is impossible.

6. In prime factorizations, squares carry even exponents (Problem 64.1). In $a^2 = 3b^2$, the exponent of $3$ is even on the left, but odd on the right (even from $b^2$, plus one). No number has two factorizations (Theorem 64.6): contradiction, and $\sqrt3$ is irrational.

7. In $a^2 = n b^2$, the argument finds a prime with contradictory exponent parity exactly when some prime appears in $n$ with an odd exponent. If every exponent of $n$ is even, $n$ 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 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: $\sqrt2$ and $-\sqrt2$ (or $\sqrt2$ and $1 - \sqrt2$) are each irrational, with rational sums $0$ and $1$. Products: $\sqrt2 \times \sqrt2 = 2$. 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.

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 $2$ 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 miss the impossible by exactly one unit — the closest whole numbers can ever come.

15. The number line must contain numbers beyond the fractions — 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.