Prove the function f:Z^+ → Z given by f(n) = (-1)^n * n is Injective(one-to-one)

Prove the function f:Z^+ → Z given by f(n) = (-1)^n * n is Injective(one-to-one)
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Prove the function f:Z^+ → Z given by f(n) = (-1)^n * n is Injective(one-to-one)

[@zawarudo4297]

Hi, first of all thanks for your effort on this site, it is very helpful! I have a doubt, why we impose n-m=0 at the exponent when we notice that the sign of n=m(-1)^(n-m) must be the same? This condition is verified if the exponent is even, so in general if n-m=2k for some k in Z. So why we deduce n-m=0 instead of n-m=2k? Thanks!

[@shamarMUNDELL]

This was simply AMAZING !

[@TheMathSorcerer]

thanks!!

[@letsexplore687]

Can u solve f:z Xz _z. Given by f(x,y) =(-1)^x y .decide whether or not injective surjective invertible

[@k.8597]

cant you just choose m and n to be distinct and then input those values in and provide cases for when both are odd or both are even and use (-1)^ odd is -1 and (-1)^ even is 1 so you eventually go from f(m)=f(n) to m = n in both cases? And if you want to consider m as odd and n as even or vice versa WLOG you couldn't consider f(m) =f(n) since f(m) = odd and f(n) = even so vacuously, m=n?

[@petter9078]

Isn't it very important to note here that (-1)^0 = 1. Which enables m = 1 * n, I think you skipped it.

[@cylin4052]

Z+ = positive integers, not including 0

[@PhillipRhodes]

How does that matter? (m-n) = 0 if m = n, and that seems to indeed be the key point. Consider that (-1)^p would be 1 for any even p, but to get p equal to, say, 2, by having m = 4 and n = 2, you'd lose the condition that m = 1 * n. It seems to me that the point that m-n = 0 is really what drives this proof. Am I missing something?

[@waatstat6789]

Thanks bro verey helpfull 🥰🥰

[@TheMathSorcerer]

No problem 👍