Arnold F. answered • 03/08/16
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For this to be injective you want to make sure two different values from Z go to different values of N [i.e. if x ≠ y then f(x)≠f(y)]. That is equivalent to saying if f(x)=f(y) then x=y (the contrapositive).
Say (|4x+1| +1)/2 = (|4y+1| +1)/2 then you must prove x=y.
But this reduces down to |4x+1| = |4y+1|.
Now do this in cases:
1. If the insides of both absolute values are positive you get 4x+1=4y+1
2. If both negative then (what?)
3. If left inside is the only negative then -4x-1= 4y+1
but this gives -x -y = 1/2 (this can't be because (what?))
4. If right inside is negative then (what?)
I left out a few steps. Can you finish it?
Let me know by commenting back.
Kyle C.
So just to be sure this means its not injective right? Do I need to do all the cases if the first case proves its not injective?
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03/08/16
Arnold F.
I believe it is injective:
1. For case 1 above to be true: 4x+1=4y+1 so x=y (this is what we needed to show for injective to hold)
2. For case 2 the same
3. Cases 3 and 4 can't happen, that is the inside of the abs val can't be negative in one and positive in the other and still be equal (try it out.)
So the result is: if f(x)=f(y) then x=y.
Let me know if this makes sense to you.
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03/08/16
Kyle C.
03/08/16