Proving That the Square Root of 2 Is Irrational
Prove that $\sqrt{2}$ is irrational, that is, that there are no positive integers $m$ and $n$ with $\left(\frac{m}{n}\right)^2 = 2$.
Hints
- Argue by contradiction: suppose $\sqrt{2} = m/n$ with $m$ and $n$ positive integers, and reduce the fraction so that $m$ and $n$ have no common factor.
- Square both sides to get $m^2 = 2n^2$. What does this say about whether $m$ is even or odd? Remember that the square of an odd number is odd.
- Write $m = 2k$ and substitute: $4k^2 = 2n^2$, so $n^2 = 2k^2$ and $n$ is even too. Now both $m$ and $n$ are even, contradicting the assumption that the fraction was in lowest terms.
Worked Solution
How to Think About It: A statement of the form "no integers satisfy this" is naturally attacked by contradiction: assume a solution exists, choose it in a canonical minimal form (lowest terms), and derive that it was not minimal after all. The equation $m^2 = 2n^2$ makes parity the natural lens.
Approach: Assume $\sqrt{2} = m/n$ in lowest terms and show both $m$ and $n$ must be even.
Formal Solution:
*Step 1 -- Assume the opposite.* Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{m}{n}$ for some positive integers $m, n$. Dividing out any common factors, we may assume $\gcd(m, n) = 1$, so $m$ and $n$ are not both even.
*Step 2 -- Square.* Squaring gives $m^2 = 2 n^2$, so $m^2$ is even.
*Step 3 -- $m$ is even.* If $m$ were odd, say $m = 2j + 1$, then $m^2 = 4j^2 + 4j + 1$ would be odd. Since $m^2$ is even, $m$ must be even: $m = 2k$ for some integer $k$.
*Step 4 -- $n$ is even.* Substituting, $(2k)^2 = 2n^2$, i.e. $4k^2 = 2n^2$, so $n^2 = 2k^2$. By the same argument as in Step 3, $n$ is even.
*Step 5 -- Contradiction.* Both $m$ and $n$ are even, so $2$ divides $\gcd(m, n)$, contradicting $\gcd(m, n) = 1$. Hence no such $m, n$ exist, and $\sqrt{2}$ is irrational.
*Step 6 -- Remarks.* The same argument with "divisible by $p$" in place of "even" proves that $\sqrt{p}$ is irrational for any prime $p$; a cleaner version compares the exponent of $2$ in the prime factorisations of $m^2$ (even) and $2n^2$ (odd), which can never be equal. Numerically the claim is consistent with the fact that $\sqrt{2} = 1.41421356\ldots$ has no repeating decimal pattern, though that observation is not a proof.
Answer: $\sqrt{2}$ is irrational. If $\sqrt{2} = m/n$ in lowest terms, then $m^2 = 2n^2$ forces $m$ even, $m = 2k$, whence $n^2 = 2k^2$ forces $n$ even as well, contradicting that $m/n$ was in lowest terms.
Intuition
The equation $m^2 = 2n^2$ forces a factor of $2$ into $m$, which then forces a factor of $2$ into $n$, and the process never stops: you can keep halving forever, which no pair of positive integers allows. Reducing the fraction to lowest terms first turns this infinite descent into a one-step contradiction. Parity (or more generally, comparing exponents of a prime on both sides) is the mechanism, and the same argument shows $\sqrt{p}$ is irrational for every prime $p$. Contradiction-by-descent is a standard proof pattern worth having at your fingertips.