Confused by proof of the irrationality of root 2: if $p^2$ is divisible by $2$, then so is $p$.

Solution 1:

The quickest proof of that fact is to note that every whole number $n$ is either even or odd.

If $n$ is even, $n=2k$ for some whole number $k$: $n^2 = 4k^2 = 2(2k^2)$ is even.

If $n$ is odd, $n=2k+1$ for some whole number $k$: $n^2 = (2k+1)^2 = 4k^2 +4k + 1 = 2(2k^2 +2k) +1$ is odd.

Therefore the square of a whole number is even if and only if that number is even.

Solution 2:

According to Euclid's lemma if $p$ is a prime and $a,b$ are integers, then $p \mid ab$ implies that $p \mid a$ or $p \mid b$.

Since $2$ is prime and it divides $p^2 = p\cdot p$ then either $2 \mid p$ or $2 \mid p$. In either case, $2 \mid p$.

Solution 3:

A simple and (maybe) less technical way to state it could be to draw the following table: $$ \begin{array}{|c|c|} \hline \times&\text{even}&\text{odd}\\ \hline \text{even}&\text{even}&\text{even}\\ \hline \text{odd}&\text{even}&\text{odd}\\ \hline \end{array} $$

As an aside, I personally prefer the proof that no non-integer fraction has non-integer powers by exploiting unique prime factorization: $$ \left(\frac{p_1\cdot p_2\cdots p_n}{q_1\cdot q_2\cdot q_m}\right)^s=\frac{p_1^s\cdot p_2^s\cdots p_n^s}{q_1^s\cdot q_2^s\cdot q_m^s} $$ since if you could not cancel any of the prime factors $p_1,p_2,...,p_n,q_1,q_2,...,q_m$ before, you still cannot do it when they appear with multiplicity $s$. This proves that among integers, only perfect squares, cubes etc. have integer square roots, cube roots, etc.

Solution 4:

Using modular arithmetic,

$$p^2\equiv p\mod2$$ because

$$p^2-p=p(p-1)\equiv0\mod2$$ as one of $p$ and $p-1$ is even.