Discrete Math Function problem

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raross

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Sep 24, 2005
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Hey if anyone could help me with this I would be sooo grateful. Im trying to grasp the idea of onto, one-to-one and bijection(both) functions.

A sample problem is: If f(x) = 2x . What is f(Z), all integers. What is f(N), all naturals. What is f(R), all real. These are 3 different problems, and im trying to figure out if they are onto, and/or one-to-one or bijection(which means both).

Any help would be a BIG HELP! Thanks!
 
In any case, f(x)=2x is one-to-one but not onto.
If f(a)=f(b), 2a=2b then a=b. Thus it is one-to-one.
In the case of Z, on odd integer is an image. So it is not onto.
 
This is what I got.

1. f(x) = 2x then f(Z) = one-to-one but not onto

2.f(x) = 2x then f(N) = one-to-one but not onto. Since this would practically be the same as all Integers. It never hits the odd values.

3. f(x) = 2x then f(R) = one-to-one AND onto. So this would be bijective.


but for all REALS, it would be bijection wouldnt it? meaning both one-to-one and onto. I feel confident in the other answers just not REAL.
 
If r is a real number then so is r/2 a real number.
Well f(r/2)=r: so isn’t f onto?
 
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