Prove the statement $\forall x\in\mathbb N (x > 1\to\exists k\in\mathbb N\exists m \in\mathbb N (m \equiv 1 (\text{mod }2) \wedge x = 2^km))$.

Question

Can someone please show me how to prove this statement? $$\forall x\in\mathbb N (x > 1\to\exists k\in\mathbb N\exists m \in\mathbb N (m \equiv 1 (\text{mod }2) \wedge x = 2^km))$$ I can only assume $m = 3$ and when $x > 1$.

Answer 1

The statement uses a lot of unnecessary notation. What it means is

Every integer $x>1$ can be expressed as an odd natural number multiplied by a power of $2$.

Can you see why this is true?

Hint: what happens if you take the largest power of $2$ that divides $x$? (Note that this might be $2^0=1$.)