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.

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.

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 ${\frac{5}{4}}$?

  • 1
    $\begingroup$ Here's how I'd do it. If $(a/b)+(c/d)=(ad+bc)/(bd)$ is an integer, then $bd$ divides $ad+bc$, so $b$ divides $ad+bc$ and $d$ divides $ad+bc$. From $b$ dividing $ad+bc$, it follows that $b$ divides $ad$, but $\gcd(a,b)=1$, so $b$ divides $d$. Similarly, from $d$ dividing $ad+bc$, it follows that $d$ divides $b$. So, $b=d$, contradiction. $\endgroup$ Jun 14, 2020 at 6:54
  • $\begingroup$ Now, in your argument, by hypothesis $a/b$ is reduced, but you use $(Nb-a)/b$ is reduced. You can't do that without justifying it. It's probably where the argument fails for $5/4$. $\endgroup$ Jun 14, 2020 at 6:57
  • $\begingroup$ @Gerry Myerson I don't recall stating that (Nb - a)/b is reduced. Clarification? $\endgroup$
    – Tauist
    Jun 14, 2020 at 7:01
  • $\begingroup$ @Tauist when you said that "the only way for two fractions to be equal are for the numerator and denominator to be the same values...", you are assuming that $(Nb-a)/b$ is reduced. You need to further prove that $(Nb-a)/b$ is indeed irreducible. $\endgroup$
    – Poypoyan
    Jun 14, 2020 at 7:06
  • 1
    $\begingroup$ If a prime $p$ divides $b$ then it divides $Nb$ and doesn't divide $a$, so it doesn't divide $Nb-a$. Therefore, $\gcd(Nb-a,b)=1$. $\endgroup$ Jun 14, 2020 at 12:39

1 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:

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

  • $\begingroup$ Could you explain what you mean with notation as well? Your second paragraph lost me. $\endgroup$
    – Tauist
    Jun 14, 2020 at 7:18
  • $\begingroup$ I added a little more, lmk if you need more explanation :) $\endgroup$
    – atul ganju
    Jun 14, 2020 at 7:30
  • $\begingroup$ Elaboration on what "multiplying top and bottom by f" means and how d = fb is reached with more notation would be greatly appreciated! $\endgroup$
    – Tauist
    Jun 14, 2020 at 7:35
  • $\begingroup$ How about now? I changed the explanation entirely, this might make more sense :) $\endgroup$
    – atul ganju
    Jun 14, 2020 at 7:42
  • 1
    $\begingroup$ I am saying that your statement can't be proven unless you ensure $N$ is an integer. If $N$ is instead only required to be a rational, then for example, in the case where $N = \frac{cb+ad}{bd}$ we can show you that the equation can be satisfied implying finally that $\frac{a}{b} + \frac{c}{d} = N$ can be satisfied. $\endgroup$
    – atul ganju
    Jun 14, 2020 at 7:51

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