2

Where x and y are in the set of all integers:

1. f(x,y) = 2x − y
2. f(x,y) = x^2 − y^2 
3. f(x,y) = x^2 − 4

Here are my answers:

1. True
2. False (counterexample would be (-2,-2) and (2,2) which result in same value
3. False (counterexample would be (-2,1) and (2,1) or (2,7), etc

Can someone tell me if these are correct and tell me how to prove if it for the functions that are one to one?

Thandor7765
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2 Answers2

2

  1. is not one to one, since $f(2,2) = f(1,0)$.

  2. and 3. are also not one to one, as you correctly showed.

5xum
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  • Ah cool, I guess that makes me avoid having to prove it. But how would I prove a function is one-to-one formally if there was one? – Thandor7765 Nov 07 '16 at 11:30
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    @Thandor7765 By assuming that $f(a)=f(b)$ and proving that therefore, $a=b$. For example, if $f(x)=x$, then it is simple to show that $f$ is injective. Take $x,y$ such that $f(x)=f(y)$. But then, since $x=f(x)$, you have $x=f(x)=f(y)$, and since $f(y)=y$, you have $x=f(x)=f(y)=y$, therefore $x=y$, so $f$ is injective. – 5xum Nov 07 '16 at 11:31
  • @Thandor7765 Naturally, things get trickier for more complicated functions, but the idea remains the same. – 5xum Nov 07 '16 at 11:33
  • For some reason I thought it would be different for functions that had multiple variables, but I guess not – Thandor7765 Nov 07 '16 at 11:34
2

  1. f(4,1)=f(3,-1)=7.

  2. f(1,1)=f(-1,-1)=0.

  3. f(1,1)=f(-1,1)=-3.

none of them is one to one.

hamam_Abdallah
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