# Is it true that odd number raised to power (natural number) will always be odd?

Is it true that odd number raised to power (natural number) will always be odd ? If this is true what is explanation behind this?

Thanks

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2

Have you learned mathematical induction?
– ervx
2 days ago

3

A number is even if one of its factors is 22. A power of an odd number, nkn^k, has only odd factors: nk=n⋅n⋯n⋅n⏟k timesn^k = \underbrace{n\cdot n\cdots n\cdot n}\limits_{k\text{ times}}.
– Bye_World
2 days ago

1

@د€r8 I have tried to expand using binomial theorem as (2r−1)n(2r-1)^n
– J. Deff
2 days ago

3

Hint: (1) a0=1a^0=1 for any odd aa; (2) an odd number times an odd number is an odd number.
– Parcly Taxel
2 days ago

1

Suppose that aka^k is even. Then 22 is a factor of aka^k in which case 22 must be a factor of a.a.
– Doug M
2 days ago

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2

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(2r−1)n=n∑k=0(nk)(2r)k(−1)n−k=(−1)n−k+n∑k=1(nk)(2r)k(−1)n−k.
(2r – 1)^n = \sum_{k = 0}^n \binom{n}{k} (2r)^k(-1)^{n-k} = (-1)^{n-k} + \sum_{k = 1}^n \binom{n}{k} (2r)^k (-1)^{n-k}.

Every term in the rightmost summation has a factor of two, so that sum is even. The remaining term of (−1)n−k(-1)^{n-k} either adds or subtracts 11, so the overall sum is odd.

1

Yeah, but using Binomial Theorem to prove this is using a piledriver to crack a peanut. As @ParclyTaxel pointed out, you easily see that the product of any two odds is odd. Proof done.
– Lubin
2 days ago

1

Certainly, but the OP mentioned having the binomial theorem in mind.
– Austin Mohr
2 days ago

This also isn’t quite the binomial theorem, as written – you’re missing the binomial coefficients?
– د€r8
yesterday

1

What an oversight! I’ve corrected the answer. Thanks for pointing it out.
– Austin Mohr
yesterday

Let a^0=1a^0=1 be the base case. Then apply the inductive step f(x)=axf(x)=ax (observe that f^n(a)=a^nf^n(a)=a^n). Clearly, this map preserves the parity, thus it must be odd.