Exercise 6.3.4 ($\sqrt[3]{10}$ is irrational (with a reasoning in base ten))

Proof.

(a)

Assume, for the sake of contradiction, that $\sqrt{10}$ is rational, so that $\sqrt{10}=h∕k$, thus $$h^2=10k^2.$$

If $h$ ends in $m$ zeros, then $h=10^m{h}^{\prime }$, where the last digit of $h^{\prime }$ is not $0$. Then $h^2=10^{2m}{h{}^{\prime }}^2$, where in base ten, the last digit of $h^{\prime }$ is not $0$. Therefore $h^2$ ends in $2m$ zeros. Similarly, if $k$ ends in $p$ zeros, then $k^2$ ends exactly in $2p$ zeros, and $10k^2$ in $2p+1$ zeros. This shows that $h^2$ ends in an even number of zeros whereas $10k^2$ ends in an odd number of zero in base ten. Hence $h^2\ne 10k^2$. This is a contraction, so $\sqrt{10}$ is irrational.

(b)

Similarly, if $\sqrt[3]{10}=h∕k$, then $h^3=10k^3$. But $h^3$ ends in $3m$ zeros in base ten, and $10k^3$ in $3p+1$ zeros. Since $3m\ne 3p+1$ (otherwise $3\mid 1$), this proves that $h^3\ne 10k^3$. This contradiction shows that $\sqrt[3]{10}$ is irrational.

(c)

We must take some caution to generalize in base $n$. For instance, if $n=75$, then $15$ does not end by $0$, but the last digit of $15^2=3\cdot 75$ in base $75$ is $0$, and $15^2$ ends in an odd number of zeros in base $75$, despite the fact that $75$ is not a perfect square. We cannot generalize the argument of part (a).

Suppose that $n$ is not a perfect square. Assume, for the sake of contradiction, that $\sqrt{n}$ is rational, so that $\sqrt{n}=h∕k,h\in \mathbb{N},k\in {\mathbb{N}}^{\text{∗}}$, and $h\wedge k=1$. Then $h^2=n{k}^2$. Therefore $k\mid h^2$, where $k\wedge h=1$, thus $k\mid 1$, and $k>0$, so $k=1$. Then $n=h^2$ is a perfect square. This contradiction shows that $\sqrt{n}$ is irrational.