Question about irrational numbers

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Discussion Overview

The discussion revolves around proving that \(\sqrt{p/q}\) is an irrational number, where \(p\) and \(q\) are distinct primes. The scope includes mathematical reasoning and proof techniques.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that assuming \(\sqrt{p/q} = a/b\) (where \(a\) and \(b\) are relatively prime) leads to a contradiction, indicating that \(\sqrt{p/q}\) must be irrational.
  • One participant questions the reasoning that \(b\) is not divisible by \(p\), seeking clarification on this point.
  • Another participant responds that if \(b\) were divisible by \(p\), it would contradict the assumption that \(GCD(a,b) = 1\).

Areas of Agreement / Disagreement

There is no consensus on the proof, as one participant expresses confusion regarding a specific step in the argument, indicating that the discussion remains unresolved.

Contextual Notes

The discussion includes assumptions about the properties of primes and relative primality, which may not be fully explored or defined in the posts.

olcyr
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Let p and q be distinct primes. Prove that [tex]\sqrt{p/q}[/tex] is a irrational number.
 
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olcyr said:
Let p and q be distinct primes. Prove that [tex]\sqrt{p/q}[/tex] is a irrational number.

It isn't a homework. I just need to prove it!

Thank you,
Olcyr.
 
It's quite easy. Assume, that [tex]\sqrt{p/q}=a/b[/tex], where a and b are relative primes, ie GCD (a,b)=1.

This is equivalent to [tex]pb^2=qa^2[/tex]. Since p and q are distinct primes, p | a^2 => p | a => The right side is divisible by p^2, and this is a contradiction, because the left side is not (because b is not divisible by p, since GCD (a,b)=1)
 
I din't understand why b isn't divisible by p.

Thank you for your answer!
 
because if b is divisible by p, than GCD (a,b) is at least p, but we assumed that it equals to 1
 
Thanks! :)
 

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