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 $x > 1$.

Can you write down, in words, what this means? That may help. If not, I can help you. — Sarvesh Ravichandran Iyer, Apr 05 '17 at 10:19

Especially Lime · Answer 1 · 2017-04-05T10:50:12.440

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 $2^0=1$.)

edited Apr 05 '17 at 10:50

answered Apr 05 '17 at 10:23

Especially Lime

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yes but I just don't know how to prove it using mathematical method, I can only prove it using examples. – jhg Apr 05 '17 at 10:41
Ok, have added a hint. – Especially Lime Apr 05 '17 at 10:51

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

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