s is irrational number, prove s/2 is also irrational

We may prove it by contradiction.

Suppose s/2 is rational, i.e. s/2 = p/q for some integers p,q and p,q are relatively prime. Thus, we have

s/2 = p/q
s = 2p/q

If q is odd, then 2p and q are relatively prime and thus s is rational which contradicts the assumption that s is irrational.

If q is even, then let q=Q*2^k where Q is odd, then s=2p/q=p/Q*2^(k-1). Note that p must be odd and p,Q are relatively prime (otherwise, p,q are not relatively prime). So, p and Q*2^(k-1) are also relatively prime. Thus, s is rational and it also contradicts the assumption.

Therefore, s/2 is also irrational.

If s is an irrational, then s cannot be weitten as p / q, where p and q are relative prime integers with q being non-zero.

So for s/2, if it is rational, then we could write it as r / t for some r and s being relative prime integers with t being non-zero.

but then s = 2r / t which is rational, which contradicts that s is an irrational, so s/2 must be irrational.