Square Root of 2 is Irrational/Proof 4
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Theorem
- $\sqrt 2$ is irrational.
Proof
Aiming for a contradiction, suppose that $\sqrt 2$ is rational.
Let $n$ be the smallest positive integer such that:
- $\sqrt 2 = \dfrac m n$
for some $m \in \Z_{>0}$
Then:
- $m = n \sqrt2 > n$
so:
- $(1): \quad m - n > 0$
We also have:
- $m = n \sqrt2 < 2 n$
so:
- $m < 2 n$
and therefore:
- $(2): \quad m - n < n$
Finally, we have
\(\ds m^2\) | \(=\) | \(\ds 2 n^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds m^2 - n m\) | \(=\) | \(\ds 2 n^2 - n m\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds m \paren {m - n}\) | \(=\) | \(\ds \paren {2 n - m} n\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {2 n - m} {m - n}\) | \(=\) | \(\ds \dfrac m n\) |
It follows that:
- $\dfrac {2 n - m} {m - n} = \sqrt2$
By $(1)$, the denominator of $\dfrac {2 n - m} {m - n}$ is positive.
By $(2)$, the denominator of $\dfrac {2 n - m} {m - n}$ is less than $n$.
We have thus written $\sqrt2$ as a fraction with a smaller denominator than $n$, which is a contradiction.
$\blacksquare$