When I was comparing proof for $\sqrt{2}$ and sum of two rational numbers, I found that the proof of two rational number did not mention anything about common factor in the ratio.

one proof I found for sum of two rational numbers:

The sum of any two rational numbers is rational.


Suppose r and s are rational numbers. [We must show that r + s is rational.] Then, by the definition of rational numbers, we have

                                        r = a/b    for some integers a and b with b ≠ 0.

                                        s = c/d   for some integers c and d with d ≠ 0.

So, by substitution, we have

                                        r + s = a/b + c/d

                                                = (ad + bc)/bd

Now, let p = ad + bc and q = bd. Then, p and q are integers [because products and sums of integers are integers and because a, b, c and d are all integers. Also, q ≠ 0 by zero product property.] Hence,

r + s = p/q , where p and q are integers and q ≠ 0.

Therefore, by definition of a rational number, (r + s) is rational. This is what was to be shown.

And this completes the proof.

Unlike $\sqrt{2}$ proof which state that because both p and q for p/q is prove to be even, therefore contradicting the premises that it is a irreducible faction, The sum proof did not mention if $\frac{ad+bc}{bd}$ might have a common factor , hence make this proof incomplete. Why is that?

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    $\begingroup$ Why is the proof incomplete? Why would it matter that your numbers have common factors? $\endgroup$ – YoTengoUnLCD Jun 10 '15 at 20:46
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    $\begingroup$ Note that your definition of rational (as used for $r,s$) does not insist that the numerator and denominator have no common factor. It just says that there are SOME numerator $a$ and nonzero denominator $b$ with $r=a/b$. In fact this choice is not unique. $\endgroup$ – vadim123 Jun 10 '15 at 20:49
  • $\begingroup$ The sum proof does not mention (ir)reducibility of the fraction because it plays no role in that proof. $\endgroup$ – Bill Dubuque Jun 10 '15 at 20:53

There is nothing incomplete in the proof. The proof shows the following statement:

If there exist such integers $a,b$ that $r=\frac ab$ and there exist such integers $c,d$ that $s=\frac cd$, then there also exist such integers $p, q$ that $r+s = \frac pq$.

Remember that this is all you need. A number $x$ is rational if and only if there exist some integers $m,n$ so that $x=\frac mn$.

The fact that there also exist two (almost) uniquely determined coprime integers $m', n'$ such that $x=\frac{m'}{n'}$ does not mean that every pair of integers that forms $x$ is coprime, only that one such pair exists.

On the other hand, for $x=\sqrt 2$, the proof shows that no such pair exists.

  • $\begingroup$ Ok, reducibility is not part of the rational number definition, and proof for square root 2 first assume there exist coprime fraction(and any other fraction that can be reduced to the coprime fraction), but that leads to contridiction that there is no coprime fraction, hence no other fraction can be based on the coprime fraction, hence no such integral fraction exist, therefore root 2 is not a rational number. Is that right? $\endgroup$ – dk000000000 Jun 12 '15 at 18:42
  • $\begingroup$ @dk000000000 Almost. The proof for $\sqrt 2$ assumes that there exists some fraction. Then, because of this, there also exists a coprime fraction. $\endgroup$ – 5xum Jun 13 '15 at 1:18

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