Verify Proof that N cannot be expressed as ${\frac{a}{b}} + {\frac{c}{d}}$ Given Certain Parameters

Question

I'm trying to prove the statement for any integer, N:

N cannot be expressed as ${\frac{a}{b}} + {\frac{c}{d}}$

Given the two parameters:

1. a, b, c, and d are all different integers which are not 0
2. The fractions, ${\frac{a}{b}}$ & $\frac{c}{d}$, are irreducible. In other words, a and b are co-prime, and c and d are co-prime.

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My current proof is a proof of contradiction and goes as follows:

Propose ${\frac{a}{b}} + {\frac{c}{d}} = N$ , we can then conclude that $N - {\frac{a}{b}} = {\frac{c}{d}}$.

The integer, N, can be rewritten as ${\frac{Nb}{b}}$ because multiplying then dividing by the same value yields the original number. If we replace n with this new, ${\frac{Nb}{b}}$, we get: $${\frac{Nb}{b}} - {\frac{a}{b}} = {\frac{c}{d}}$$ Since the terms share a common denominator, this can be simplified to: $${\frac{Nb - a}{b}} = {\frac{c}{d}} $$

Due to parameter 2, the only way for these two fractions to be equal are for the numerator and denominator to be the same values since there are no reducible fractions allowed.

This would mean that b and d would have to be equal in order for the equation above to hold true, yet parameter 1 states b and d cannot be equal. This is a contradiction which thus proves that ${\frac{a}{b}} + {\frac{c}{d}} = N$ is false, or alternatively: $${\frac{a}{b}} + {\frac{c}{d}} \neq N$$ is true.

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My Question:

Is this proof valid? Is it true that N cannot be expressed as ${\frac{a}{b}} + {\frac{c}{d}}$?

And as a bonus question:

The fraction, ${\frac{5}{4}}$, can be written as ${\frac{a}{b}} + {\frac{c}{d}}$ given the parameters mentioned (${\frac{1}{2}} + {\frac{3}{4}}$).

If my proof is indeed valid for all integers, n, why is not also valid for fractions which are not whole such as ${\frac{5}{4}}$?

Answer 1

You are correct in your proof for when N is an integer! Now, it is important to emphasize that the new fraction $\frac{Nb-a}{b}$ is irreducible. To do so, we can use Euclidean's Algorithm to show the gcd of the numerator and denominator is 1:

$gcd(Nb-a,b) = gcd((N-1)b-a,b) = ... = gcd(a,b) = 1$

this is true because $N$ is an integer so $Nb$ is an integral multiple of $b$. However, this only works because $N$ is an integer. If $N$ is a non-integral rational (i.e. $N = \frac{e}{f}$), then if $N = \frac{cb+ad}{bd}$ we get:

$\frac{Nb-a}{b} = \frac{\frac{cb+ad}{bd}b-a}{b} = \frac{c}{d}$