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major_maths
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1. Assume that f : A -> B and g : B -> C and that f is not one
to one. Prove that g(f) is not one to one.
2. one-to-one: x1 != x2 and f(x1) != f(x2)
not one-to-one: f(x1)=f(x2) and x1 != x2
3. The start of the proof should go as follows:
Assume f: A->B, g: B -> C, f is not 1-1, g is 1-1.
Since f is not 1-1, there exists an x1 and an x2 in the dom(f) such that f(x1) = f(x2) and x1 != x2.
Since g is 1-1, for every x1 and x2 in dom(f), if x1 != x2 then g(x1) != g(x2).
And that's as far as I got. I'm not sure if I should assume g is 1-1 since that's not explicitly in the instructions, but I don't thing g(f) could be 1-1 if g wasn't 1-1. I tried plugging f(x1) into g but I couldn't find a logical way to get to the conclusion by doing that.
to one. Prove that g(f) is not one to one.
2. one-to-one: x1 != x2 and f(x1) != f(x2)
not one-to-one: f(x1)=f(x2) and x1 != x2
3. The start of the proof should go as follows:
Assume f: A->B, g: B -> C, f is not 1-1, g is 1-1.
Since f is not 1-1, there exists an x1 and an x2 in the dom(f) such that f(x1) = f(x2) and x1 != x2.
Since g is 1-1, for every x1 and x2 in dom(f), if x1 != x2 then g(x1) != g(x2).
And that's as far as I got. I'm not sure if I should assume g is 1-1 since that's not explicitly in the instructions, but I don't thing g(f) could be 1-1 if g wasn't 1-1. I tried plugging f(x1) into g but I couldn't find a logical way to get to the conclusion by doing that.