Attention Math Nerds

from Elemental Forums

Define f : ℤ_n -> ℤ_n by f(x) = x^2

For which n ∈ ℕ is f injective?

 

Nobody in the class got it right. He told us to ponder it. I know x ≡ -x must be true, and this means n = 1 or n = 2 but I can't prove it. So x ≡ -x leads to 2x ≡ 0 mod n, which leads to 2x = kn + 0, from here if I could just prove k is odd I'd be done, but I can't.

 

So, any clues or insights on how to show k is odd or any completely different way to solve this problem?

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44,737 views 12 replies
Reply #1 from Elemental Forums Top

I don't want to do your homework for you, so I'll just suggest that you drop the fancy algebra and look for some concrete counterexamples.  With any luck, the answer will make itself obvious to you, and then you can prove it algebraically. :)

Reply #2 from Elemental Forums Top

That is exactly one of the main reasons I'm not going to university. Math is too hard >:(

 

Like what the hell is = with the extra - ?? o_O

Then <letter>_<letter>   what..?

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Reply #3 from Elemental Forums Top

This is why I hate math and excel in statistics. Math is for someone else :)

Reply #4 from WinCustomize Forums Top

I gave up on math as soon as they started including letters! :P

Reply #5 from WinCustomize Forums Top

I gave up on any form of math that does not involve getting the correct change at a cash register a long time ago. ;P

Problem being, most people cannot even take 37 away from 50 for instance without a calculator nowadays. Sad, very sad.

Reply #6 from WinCustomize Forums Top

Problem being, most people cannot even take 37 away from 50 for instance without a calculator nowadays. Sad, very sad.
End of quote

LOL.. isn't it incredible the looks you get from a cashier these days when you hand them the odd cents to get back a full bill.

COUNTING change back is a lost art, as with many other things today.

Reply #7 from WinCustomize Forums Top

math is for ugly people  :grin:

Reply #8 from WinCustomize Forums Top

Er............... 3 o'clock. B[]

Reply #9 from Elemental Forums Top

Math is fine.  It's posting algebraic equations on an internet site that I've given up on.

 

Happy Pi Day, by the way.

Reply #10 from Elemental Forums Top

Firstly i'm not clear what the context is. So I'm assuming you're thinking of the additive group (Z/n) and you want to see when this is an injective homomorphism.

 

Okay, homomorphism first:

you want f(a + b) = f(a) + f(b),

so 2(a + b) = 2a + 2b. So that's fine :-)

and f(0) = 2x0 = 0.

 

If it's injective the kernel must be trivial (Otherwise, suppose x non-trivial in the kernel. Then f(x) = f(0), so not injective).

 

Also trivial kernel => injective here. Prove this by contra-positive:

f not injective. So f(a) = f(b), for some a not equal b, say.

then f(a - b) = f(a) - f(b) = f(a)  -f(a) = 0. So a-b is in the kernel. Therefore the kernel is non-trivial.

 

So you're looking for all n such that the kernel is trivial. If you want to rule out a specific n, just find something non-trivial in the kernel and you're done.

 

I guess you've probably solved it by now but, if not, there's a bit of a nudge :-).

Reply #11 from WinCustomize Forums Top

Additives are bad for you! I stay away from the homomorphism stuff...to political. And injecting kernels is bad for you...unless of course you take the husk off first...the husk hinders the a-b in the kernel. And never rule out the n! Long live the N!;)

Reply #12 from Elemental Forums Top

Computer kernals have husks?   Does that mean some computers might become husky?