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