Thread: Fraction in Lowest terms

Prove is a fraction in its lowest terms

Rewriting


The numerator is a product, so we shall consider them separately


Let

. Numerator and denominator share no common factors

Consider

This also has no common factors

Hence

has no common factors
QED

Is this proof valid?

It believe it is correct, but whether it is acceptable to an instructor depends on your course. You may have studied an algorithm to prove similar claims, and then you need to go over the steps without justifying them. However, for this to look like a proof to an outsider, it should include not only computational steps, but also invocation of some lemmas that prove that the computations indeed show what you claim they do.

Originally Posted by I-Think

Rewriting


The numerator is a product, so we shall consider them separately

Here one should say that n and n^2 + 1 are irreducible (why?) and so, if some polynomial divides n(n^2+2), then it is either n or n^2+2, using something like Euclid's lemma for polynomials. Therefore, it is sufficient to check whether the denominator is divisible by these two polynomials. Then one should say a word or two why does not divide and does not divide .

I think that's correct. Here is another way to approach it:

Write . Since is irriducible over , either or must be a factor of
for the fraction not to be on the lowest form. It's obvious that isn't a factor of
. For to be a factor, we must have , for some
. Expanding the RHS we have .
Comparing the coefficients we have (from the coefficient of [LaTeX ERROR: Convert failed] ) , and (from the coefficient of )
, so . But (from the coefficient of ) , so . A contradiction.