Irrational Number/Examples/Cube Root of 2

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Example of Irrational Number

$\sqrt [3] 2$ is irrational.


Proof

Aiming for a contradiction, suppose $\sqrt [3] 2 = \dfrac m n$ for integers $m$ and $n$ such that:

$m \perp n$

where $\perp$ denotes coprimality.

Then:

$m^3 = 2 n^2$

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

Hence:

$m = 2 k$

for some $k \in \Z$.

Then:

$8 k^2 = 2 n^2$

and so $2 \divides n$.

But then we have $2 \divides m$ and $2 \divides n$

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

From this contradiction the result follows.

$\blacksquare$


Sources