Proving One-to-One Functions: An Introduction to Counterexamples

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I am being asked to prove the following:
f: Z --> Z by f(n) = |2n-1| is not one to one and g:{n element of Z|n>=10} --> Z with g(n) = |2n-1| is one to one. Can anyone help me get started on this? The example done in class involved substituting in and finding out if the values were equal. Maybe it's the absolute value bars, not sure how to prove this in this instance.

Incidentally, can you have a discussion about functions without having discussed cartesian products of sets and also relations?
 
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A graph should help you see what's going on -- even if you can't use it in your formal solution.
 
see when function is one to one it means that whenever you have f(x1)=f(x2) it implies x1=x2...

so to prove that a function is NOT one to one, you have to come up with a counterexample where f(x1)=f(x2) is true and x1=x2 is false.
now for function f , what can you tell about the ordered pairs (0,1) and (1,1) ?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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