MHB Can Every Integer Be Decomposed into an Odd Integer and a Power of 2?

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Every natural number can be expressed as a product of an odd integer and a non-negative integer power of 2. For odd integers, this is straightforward as they can be represented as n multiplied by 2^0. Even integers, including zero, can be factored into 2 raised to a positive integer and another integer. The discussion emphasizes the importance of including 1 as an odd factor when considering powers of 2. A proof can be constructed using repeated division by 2 and strong induction to validate this claim.
KOO
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Prove that every n E N can be written as a product of odd integer and a non-negative integer power of 2.

For instance: 36 = 22 * 9
 
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KOO said:
Prove that every n E N can be written as a product of odd integer and a non-negative integer power of 2.

For instance: 36 = 22 * 9

If n is odd, we can write it as \displaystyle \begin{align*} n = n \cdot 2^0 \end{align*}.

If n is even (including 0), it must have a factor of 2, so we can write it as \displaystyle \begin{align*} n = 2^1 k \end{align*}.

Q.E.D.
 
Prove It said:
If n is odd, we can write it as \displaystyle \begin{align*} n = n \cdot 2^0 \end{align*}.

If n is even (including 0), it must have a factor of 2, so we can write it as \displaystyle \begin{align*} n = 2^1 k \end{align*}.
This is not enough to prove the required claim.
 
KOO said:
Prove that every n E N can be written as a product of odd integer and a non-negative integer power of 2.

For instance: 36 = 22 * 9
What about powers of 2? 2^k (k being a positive integer) has no odd factors, unless you want to include 1.

-Dan
 
topsquark said:
What about powers of 2? 2^k (k being a positive integer) has no odd factors, unless you want to include 1.
Well, we want to include 1. Thus, if $k$ is a nonnegative integer, then $2^k=1\cdot2^k$: here 1 is an odd integer and $2^k$ is a non-negative integer power of 2, so this factorization satisfies the problem statement.

Obviously, a proof of this fact uses repeated division by 2. It can be made precise using strong induction.

There was https://driven2services.com/staging/mh/index.php?posts/33828/ here about MathStackExchange (MSE), and the format that encourages dialogue was mentioned as a feature that distinguishes this forum from MSE. So I suggest that the OP write his/her reaction to what has been said so far and also the topic that this problem is supposed to teach (such as strong induction, direct proofs, or divisibility).
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

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