Square Root of 2 Is Irrational

Theorem

$\sqrt 2$ is irrational.

Classic Proof

First we note that, from Parity of Integer equals Parity of its Square, if a number is even, its square root, if an integer, is also even.

Thus it follows that:

$(A) \qquad 2 \mathop \backslash p^2 \implies 2 \mathop \backslash p$

where $2 \mathop \backslash p$ indicates that $2$ is a divisor of $p$.

Now, assume that $\sqrt 2$ is rational.

So $\displaystyle \sqrt 2 = \frac p q$ for some $p, q \in \Z$ and $\gcd \left({p, q}\right) = 1$.

Squaring both sides yields:

$\displaystyle 2 = \frac {p^2} {q^2} \iff p^2 = 2q^2$

Therefore, $2 \mathop \backslash p^2 \implies 2 \mathop \backslash p$ (see $(A)$ above).

That is, $p$ is an even integer.

So $p = 2k$ for some $k \in \Z$.

Thus:

$2 q^2 = p^2 = \left({2 k}\right)^2 = 4 k^2 \implies q^2 = 2k^2$

so by the same reasoning

$2 \mathop \backslash q^2 \implies 2 \mathop \backslash q$

This contradicts our assumption that $\gcd \left({p, q}\right) = 1$, since $2 \mathop \backslash p, q$.

Therefore, from Proof by Contradiction, $\sqrt 2$ cannot be rational.

$\blacksquare$

Proof 2

This is a special case of the result that the square root of any prime is irrational.

$\blacksquare$

Proof 3

Seeking a contradiction, assume that $\sqrt 2$ is rational.

Then $\sqrt 2 = \dfrac p q$ for some $p,q \in \Z_{>0}$

Consider the quantity $\left({\sqrt 2 - 1}\right)$:

	\(\displaystyle \)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle <\)	\(\sqrt 2\)	\(\displaystyle <\)	\(\displaystyle \)	\(\displaystyle 2\)	\(\displaystyle \)	Ordering of Squares in Reals
	\(\displaystyle \implies\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle <\)	\(\sqrt 2 - 1\)	\(\displaystyle <\)	\(\displaystyle \)	\(\displaystyle 1\)	\(\displaystyle \)

Now, observe that for any $n \in \Z_{>0}$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\sqrt 2 - 1}\right)^n\)	\(=\)	\(\displaystyle \sum_{k \mathop = 0}^n \binom n k \left({\sqrt 2}\right)^k \left({-1}\right)^{n-k}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Binomial Theorem
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{\substack{0 \mathop \le k \mathop \le n \\ k \, \text{even} } } \binom n k 2^{k/2} \left({-1}\right)^{n-k} + \sqrt 2 \sum_{\substack{0 \mathop \le k \mathop \le n \\ k \, \text{odd} } } \binom n k 2^{\left({k-1}\right)/2} \left({-1}\right)^{n-k}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a_n + b_n \sqrt 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	for some integers $a_n, b_n$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a_n + b_n \left({\frac p q}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	recall the assumption that $\sqrt 2 = \dfrac p q$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {a_n q + b_n p} q\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\ge\)	\(\displaystyle \frac 1 q\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	since the numerator is an integer and $\sqrt 2 - 1 > 0$

By Power of a Number Less Than One:

$\displaystyle \lim_{n \to \infty} \left({\sqrt 2 - 1}\right)^n = 0$

where $\lim$ denotes limit.

Recall the definition of $a_n$ and $b_n$.

By Lower and Upper Bounds for Sequences:

$0 = \displaystyle \lim_{n \to \infty} \frac {a_n q + b_n p} q \ge \frac 1 q$

a contradiction.

$\blacksquare$

Decimal Expansion

The decimal expansion of $\sqrt 2$ starts:

$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$

This sequence is A002193 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $1.3$