Proving Set Equality Using Algebraic Manipulation

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Prove C = G

C = {x + 7: x ∈ N} and G = {x : x ∈ N and x > 7}

I can come up with a logical answer to this, but I can't come up with an algebraic answer, and I'm not sure if that matters, I just want to see if there is a more proper way to do this than what I came up with, which is:

Let m ∈ N and x ∈ N, then m = x + 7 is such that m > 7 which is the definition of G. Because they share the same definition C ⊆ G and G ⊆ C and C = G.

And here's another one, same situation, I can do it logically but not algebraically:
{3x : x ∈ N} ∩ {x : x ∈ N} = {3x + 21 : x ∈ N}
 
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It seems to me like you can use induction :-).
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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