Exercises — Irrational numbers — √2, π, e — proof that √2 is irrational

A number is rational if it can be written as one integer over another.

• $\frac{7}{4}$ — already a fraction → rational.
• $0.\overline{6}=0.666\cdots =\frac{2}{3}$ (a repeating decimal is always rational, see Decimal Expansions) → rational.
• $\sqrt{9}=3=\frac{3}{1}$ → rational.
• $-3=\frac{-3}{1}$ → rational.
• $\sqrt{2}=1.41421356\dots$ never terminates, never repeats → irrational.

Answer: only $\sqrt{2}$.

It guarantees $\gcd (p,q)=1$ — $p$ and $q$ share no common factor. The proof later shows both are even (a common factor of $2$), which contradicts exactly this. That collision is the whole point.

Square both sides: $2=\frac{49}{25}$, i.e. $50=49$. False. This is the numerical echo of Step 3 of the parent proof ($2q^2=p^2$): here $2\cdot 5^2=50$ should equal $7^2=49$, and it doesn't. A specific fraction dies instantly; the general proof kills all of them at once.

Use Proof by Contradiction. Suppose $p$ is odd. Then $p=2k+1$ for some integer $k$. Square it: $p^2=(2k+1{)}^2=4k^2+4k+1=2\underset{\text{integer}}{\underbrace{(2k^2+2k)}}+1.$ That is "$2\times (\text{integer})+1$" — an odd number. So an odd $p$ gives an odd $p^2$. Contrapositive: if $p^2$ is even, $p$ cannot be odd, so $p$ is even. $\square$

Divide both sides by $2$: $q^2=2m^2$. So $q^2$ is even, hence (by Problem 2.2) $q$ is even. Now $p$ and $q$ are both even → both divisible by $2$ → $\gcd (p,q)\ge 2$, contradicting lowest terms. $\square$

Assume $\sqrt{3}=\frac{p}{q}$ in lowest terms. Square: $3=\frac{p^2}{q^2}$, so $p^2=3q^2$. Key fact used: $3$ is prime, so if $3\mid p^2$ then $3\mid p$. (An analogue of the even/odd step: a prime dividing a square must divide the base — see Prime Factorization.) Then $p=3m$. Substitute: $(3m{)}^2=3q^2\Rightarrow 9m^2=3q^2\Rightarrow q^2=3m^2$. So $3\mid q^2\Rightarrow 3\mid q$. Both $p,q$ divisible by $3$ → common factor → contradicts $\gcd (p,q)=1$. So $\sqrt{3}$ is irrational. $\square$

Look at the equation $p^2=N{q}^2$ where $N$ is the number under the root. Every square number has every prime appearing an even number of times in its factorization (because you doubled each exponent by squaring).

• For $N=2$: $p^2=2q^2$. The left side has the prime $2$ an even number of times; the right side has it an odd number of times ($2$ once, times the even count inside $q^2$). Even $\ne$ odd → impossible → contradiction. Good.
• For $N=4=2^2$: $p^2=4q^2$. Now $4$ contributes an even count of $2$'s, so both sides can match. $p=2q$ works, giving $\sqrt{4}=2$ — no contradiction.

Rule: $\sqrt{N}$ is irrational $⟺$ $N$ is not a perfect square, because only then does some prime end up on the two sides an odd-versus-even number of times.

$\sqrt{18}=\sqrt{9\cdot 2}=3\sqrt{2}$. Since $\sqrt{2}$ is irrational and $3$ is a nonzero rational, $3\sqrt{2}$ is irrational. (If $3\sqrt{2}$ were rational, dividing by $3$ would make $\sqrt{2}$ rational — contradiction.) Or directly: $18=2\cdot 3^2$ is not a perfect square (the prime $2$ appears an odd number of times), so by the rule in 3.2, $\sqrt{18}$ is irrational. $\square$

For a square of side $q$, the diagonal $d$ satisfies (Pythagoras) $d^2=q^2+q^2=2q^2.$ We assumed the diagonal is the integer $p$, so $p^2=2q^2$ — exactly the parent equation. Running the parity argument: $p$ even $\Rightarrow$ $q$ even $\Rightarrow$ both even $\Rightarrow$ common factor $2$, contradicting lowest terms. Geometric meaning: you can never build a square whose side and diagonal are both whole numbers with no shared unit — the diagonal and side are incommensurable. This is precisely what shattered the Greek belief that all lengths are ratios. $\square$

Suppose $1+\sqrt{2}$ were rational, call it $r$. Then $\sqrt{2}=r-1$. The difference of two rationals is rational, so $\sqrt{2}$ would be rational — contradicting what we proved. Hence $1+\sqrt{2}$ is irrational. $\square$ (Same trick shows $a+b\sqrt{2}$ is irrational for any rationals $a$, $b\ne 0$.)

$\sqrt{2}$ is a root of $x^2-2=0$, a polynomial with integer coefficients. By definition it is algebraic (see Algebraic vs Transcendental Numbers). So $\sqrt{2}$ is irrational and algebraic — irrationality and transcendence are different questions. ($\pi$ and $e$ are the stronger case: irrational and transcendental.)

Start with any positive integers $p,q$ (not assuming lowest terms) satisfying $p^2=2q^2$. The parity argument gives $p=2m$ and $q=2n$. Substituting $p=2m$: $(2m{)}^2=2q^2\Rightarrow 4m^2=2q^2\Rightarrow q^2=2m^2.$ And with $q=2n$: $(2n{)}^2=2m^2\Rightarrow n^2...$ — cleaner to say it plainly: the pair $(m,n)=\left(\frac{p}{2},\frac{q}{2}\right)$ also satisfies $m^2=2n^2$, and $0<m<p$, $0<n<q$. So any solution begets a strictly smaller solution. Repeat and you get an infinite strictly-decreasing chain of positive integers $p>m>\cdots >0,$ which is impossible — positive integers cannot decrease forever (there is no positive integer below $1$ that is still a positive integer). Therefore no positive-integer solution to $p^2=2q^2$ exists, so $\sqrt{2}$ is not a ratio of integers. $\square$ This version needs no GCD assumption at all: the descent itself is the contradiction.

Suppose $p^2=2q^2$ for positive integers $p,q$. Count how many times the prime $2$ divides each side.

• Any integer $n$ contains the prime $2$ some number of times, say $e$; then $n^2$ contains it $2e$ times — an even number.
• So $p^2$ has an even number of factors of $2$, and $q^2$ has an even number too.
• Then $2q^2$ has an odd number of factors of $2$ (one extra $2$ out front).
• But $p^2$ (even count) $=2q^2$ (odd count) forces even $=$ odd — impossible.

Hence no such $p,q$ exist and $\sqrt{2}$ is irrational. $\square$ This is the same "odd exponent" idea from Problem 3.2, stated as a one-line count.

Suppose ${\log }_{2}3=\frac{p}{q}$ with positive integers $p,q$. By definition of logarithm, $2^{p/q}=3$, so raising both sides to the $q$: $2^p=3^q.$ The left side is a power of the prime $2$ (only $2$'s in its factorization); the right side is a power of the prime $3$ (only $3$'s). By unique Prime Factorization, the only way $2^p=3^q$ is $p=q=0$, but $q\ne 0$. Contradiction. So ${\log }_{2}3$ is irrational. $\square$

Why is $\sqrt{N}$ irrational exactly when $N$ is not a perfect square? ::: Because $p^2=N{q}^2$ forces some prime of $N$ to appear an odd number of times on one side and an even number on the other — impossible unless $N$ is itself a perfect square (all exponents even).