Irrational Number/Examples/Square Root of 3

Example of Irrational Number

$\sqrt 3$ is irrational.

Proof

Aiming for a contradiction, suppose $\sqrt 3 = \dfrac m n$ for integers $m$ and $n$ such that:

$m \perp n$

where $\perp$ denotes coprimality.

Then:

$m^2 = 3 n^2$

Thus $3 \divides m^2$ and so $3 \divides m$.

Hence:

$m = 3 k$

for some $k \in \Z$.

Then:

$9 k^2 = 3 n^2$

and so $3 \divides n$.

But then we have $3 \divides m$ and $3 \divides n$

Hence $m$ and $n$ are not coprime after all.

From this contradiction the result follows.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.20 \ (6)$