Prove the function f:Z x Z → Z given by f(m,n) = m + n - 3 is Onto(Surjective)

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Prove the function f:Z x Z → Z given by f(m,n) = m + n - 3 is Onto(Surjective)

[@chazshot5353]

Thank you for the video. :)

[@ZenathD]

Instead of setting m=0, what about isolating both n and m so that we get: f(y-n+3, y-m+3) = (y-n+3) + (y-m+3) - 3 = 2y-m-n+3 , we then note that m=y-n+3 so that we have: 2y -(y-n+3)-n+3 =2y-y+n-3-n+3 = y would that be a stronger proof?

[@jgc9199]

Can you prove onto by showing that the range and codomain are equal to each other? I read something like that on a book but havent seen anyone proving it like that.

[@TheMathSorcerer]

Solid question, and yes you can, but there is a reason why most people don't do it. Say you have f:A->B, then we say it's onto if f(A) = B. Ok now f(A) is a subset of B by definition, here f(A) = {f(x) : x in A} and all the elements of the form f(x) are in B by definition of the funciton, so really you only need to show B is a subset of f(A) in orde r to show f(A) = B. But check this out: f is onto if for all y in B there is x in A such that y = f(x), in other words, for all y in B, we have that y is in f(A), so B is a subset of f(A), so by using the "straight up definition" you are basically showing the one inclusion that is missing. I hope that made sense. Your question was a good one, sorry for the ranty reply!!

[@jgc9199]

@@TheMathSorcerer Thanks for the response, it did make sense. I just never really rely on that when I proved onto on because I felt like you needed more than just that.

[@alissapurplebunnies5955]

Thank you!!

[@TheMathSorcerer]

np!!

[@asr84]

Thank you very much, it does make sense! And now I can clearly see through this problem.

[@TheMathSorcerer]

Awesome !

[@jong-pingkim3840]

This video is so good, I wonder why no one comments this?????

[@jean28sand]

why not (y+3,0) is a pre-image?

[@Onnethox]

Thanks. Do you have an example where instead of it being from Z x Z → Z, its instead Z → Z x Z?

[@TheMathSorcerer]

nope sorry! I should make more of these!!!!

[@marioluigi9599]

@@TheMathSorcerer Do you have a video on what that even means? Like what does the arrow mean? f colon Z arrow to Z ???

[@TheMathSorcerer]

Umm hmm yes I think I will find it for you!!

[@TheMathSorcerer]

kzbin.info/www/bejne/gWrWmoaerr53bpo

[@marioluigi9599]

@@TheMathSorcerer oh great! So the first one is the domain, which is the input to the function... ...and then the arrow points at the output codomain, which can also contain additional numbers that aren't outputs ☺️👍 Thanks, now I finally know what that's even supposed to be lol

[@mahmoudmroweh7730]

Nice vedio i have a question for you if you can help me to solve it Let f:R×R--->R defind by f(m,n)=m+n shwo that f is not injective .

[@bellamatela4273]

Muy bueno. Thank you.

[@TheMathSorcerer]

de nada!

[@krisscage]

My function is: f: Z x Z -> Z, f(x) = m + n +1 Is this same way with yours?

[@Xhhdebah]

yes

[@serinacat4781]

sorry I now realized 0 is integer. good video👍

[@TheMathSorcerer]

:)

[@serinacat4781]

I am confused because I think 0 is not integer