- #1
2^Oscar
- 45
- 0
Hi guys,
I've been reading a chapter from a book and I've been attempting a question which uses the binomial theorem to prove Fermat's Little Theorem. The question goes as follows:
Let p be a prime number:
i) Show that if r, s are positive integers such that r divides s, p divides r and p does not divide s, then p divides [tex]\frac{r}{s}[/tex].
ii) Deduce that p divides the binomial coefficient [itex]\displaystyle \binom{p}{k}[/itex] for any k such that [tex]1 \leq k \leq p-1[/tex].
iii) Now use the binomial theorem to prove by induction on n that p divides n[tex]^{p}[/tex] - n for all positive integers n. Hence deduce Fermat's Little Theorem.I can handle the first two parts of the question, but I think I may not have showed them in a way which leads onto being able to prove the third part. For i) I said that since r divides s I can express s as some multiple of r, which gets me r in both the numerator and the denominator of the fraction and thus since p divides r p must divide the fraction. I did a similar approach for the second part except using the expanded [tex]\frac{p!}{k!(p-k)!}[/tex] form.
Could someone please help me finish off the rest of the question? I'm really stumped...Cheers,
Oscar
I've been reading a chapter from a book and I've been attempting a question which uses the binomial theorem to prove Fermat's Little Theorem. The question goes as follows:
Let p be a prime number:
i) Show that if r, s are positive integers such that r divides s, p divides r and p does not divide s, then p divides [tex]\frac{r}{s}[/tex].
ii) Deduce that p divides the binomial coefficient [itex]\displaystyle \binom{p}{k}[/itex] for any k such that [tex]1 \leq k \leq p-1[/tex].
iii) Now use the binomial theorem to prove by induction on n that p divides n[tex]^{p}[/tex] - n for all positive integers n. Hence deduce Fermat's Little Theorem.I can handle the first two parts of the question, but I think I may not have showed them in a way which leads onto being able to prove the third part. For i) I said that since r divides s I can express s as some multiple of r, which gets me r in both the numerator and the denominator of the fraction and thus since p divides r p must divide the fraction. I did a similar approach for the second part except using the expanded [tex]\frac{p!}{k!(p-k)!}[/tex] form.
Could someone please help me finish off the rest of the question? I'm really stumped...Cheers,
Oscar