Does Defining ##g(y)## as ##h(y)^n## Validate the Statement?

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Vibhukanishk
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What to say about this?
Handwritten proof: If K^n = g(y), then K = h(y) for all (K, y) in Q+ and n in Z+. Since K^n = g(y) and K in Q+, g(y)^(1/n) in Q+, so g(y) = {h(y)}^n, therefore K^n = {h(y)}^n, so K = h(y). q.e.d.

Is the logic used in the solution supports the statement?
 
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There is quite a bit of information missing. E.g., there is no quantifier on ##h(y)## in the initial statement.
 
TeethWhitener said:
No. You just asserted that ##g(y)=h(y)^n## with no justification.
Handwritten-style typeset text of a proof argument: 'what it means to say in g(y)=(h(y))^n is that since (g(y))^(1/n) in Q therefore g(y) somewhere must be reducible to form (h(y))^n so (g(y))^(1/n) in Q. As K in Q implies K^n in Q implies g(y) in Q at y=a in Q. Doesn't the logic explanatory?
 
If you define ##g(y)## as ##h(y)^n##, then of course it's true, but there's also nothing to prove; it's all definitions.