anonymous
  • anonymous
How can I use Euclid's lemma to prove that if \(p\) is a prime number with \(1\leq k
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Euclid's lemma states that if \(p|ab\), then \(p|a\) and/or \(p|b\).
anonymous
  • anonymous
Do you want me to show you what I have so far?
asnaseer
  • asnaseer
surely p! (in the numerator) is ALWAYS divisible by p?

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anonymous
  • anonymous
Proof:\[\frac{p!}{k!(p-k)!}=\frac{1\cdot2\cdot...\cdot(p-1)\cdot p}{1\cdot2\cdot...\cdot k\cdot[1\cdot2\cdot...\cdot(p-k-1)\cdot(p-k)]}.\]However, since \(0
asnaseer
  • asnaseer
yes - that is what I was trying to say in the earlier statement.
anonymous
  • anonymous
But my mentor is asking me to use Euclid's lemma. :/
asnaseer
  • asnaseer
ok - let me think about it...
asnaseer
  • asnaseer
ok, maybe something like this: if p is prime, then none of the integers in k! or (p-k)! can be factors of p therefore all the integers in k! and (p-k)! must be relatively prime to p hmmm - I can't see quite how to use Euclids Lemma here. The lemma itself states that: For two integers a and b, suppose d|ab. Then if d is relatively prime to a, then d divides b.
asnaseer
  • asnaseer
ah - hang on, maybe this will work:\[\frac{p!}{k!(p-k)!}=p\times\frac{(p-1)!}{k!(p-k)!}\]now we know that k!(p-k)! is relatively prime to p, therefore k!(p-k)! divides (p-1)!
anonymous
  • anonymous
Oh! I didn't think of separating the fraction like that.
asnaseer
  • asnaseer
yes I think that works with one change in the statement: now we know that all the integers in k!(p-k)! are relatively prime to p, therefore some of the integers in k!(p-k)! must divide (p-1)!
anonymous
  • anonymous
I can't thank you enough for your help, but I'm a little confused: doesn't Euclid's lemma first assume that d|ab? That's what we're trying to show. :/
asnaseer
  • asnaseer
hmmm.. - good point - needs more thought...
anonymous
  • anonymous
I don't know. I mean, maybe my professor skimmed through this problem because wouldn't simply showing that\[\frac{p!}{k!(p-k)!}=p\cdot\frac{(p-1)!}{k!(p-k)!}\]immediately imply that the whole thing is divisible by \(p\)?
asnaseer
  • asnaseer
yes - you are right. maybe there is something he forgot to state about the problem.

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