proof for all n {N*} there exist k {N} and m {N*} with m being odd, for n =(2^k)m

[bouhrassa]

proof that forall n {N*}, there exist k {N} and m {N*} with m being odd, for n =(2^k)m

For the odd numbers, if k = 0 and n = m, then we got n = 1*n = n

For the even numbers, I don't get at all how to prove it.

Thank you for your help

 

[Ishuda]

proof that forall n {N*}, there exist k {N} and m {N*} with m being odd, for n =(2^k)m

For the odd numbers, if k = 0 and n = m, then we got n = 1*n = n

For the even numbers, I don't get at all how to prove it.

Thank you for your help

To be picky, shouldn't that be: If n is odd, let k=0 and m=n, the 2⁰ * m=1*n=n.

What is the definition of an even number? What happens if k = 1.

 

[stapel]

proof that forall n {N*}, there exist k {N} and m {N*} with m being odd, for n =(2^k)m

For the odd numbers, if k = 0 and n = m, then we got n = 1*n = n

For the even numbers, I don't get at all how to prove it.

Characters appear to be missing, and one definition may need to be specified. I think the following is what you mean:

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Let \(\displaystyle \,\left\{\mathbb{N}^*\right\}\,\) be the set of strictly positive natural numbers, starting with the number 1. Let \(\displaystyle \,\left\{\mathbb{N}\right\}\,\) be the set of non-negative natural numbers, starting with the number 0.

Prove that, \(\displaystyle \, \forall\, n\, \in\, \left\{\mathbb{N}^*\right\},\, \exists\, k\, \in\, \left\{\mathbb{N}\right\},\, m\, \in\, \left\{\mathbb{N}^*\right\},\, m\, \mbox{ odd, }\,\) such that \(\displaystyle \, n\, =\, (2^k)m\)

My proof:

For n odd, let k = 0 and let n = m . Then we get n = (2^0) m = (1)m = 1n = n.

However, for even numbers, I don't know how to do the proof.

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Is the above correct? Thank you!