Thread: Binomial Coefficients

Show that if p is a prime and k is an integer such that , then p divides

Originally Posted by EquinoX

Show that if p is a prime and k is an integer such that , then p divides

Here's the basic idea. You can put this more exact language as you please.


Since p is prime, no number nor any number divides p. Thus no matter what cancellations we have in simplifying the factorials a p remains in the numerator. etc, etc.

-Dan

Okay I kind of get your idea here, because p here is prime in the numerator, but nothing in the denominator can divide p because p is a prime, therefore p still exists after the counting and because p still exists therefore it can be divided by p? right

Originally Posted by EquinoX

Okay I kind of get your idea here, because p here is prime in the numerator, but nothing in the denominator can divide p because p is a prime, therefore p still exists after the counting and because p still exists therefore it can be divided by p? right

That is correct.

-Dan

Originally Posted by EquinoX

Show that if p is a prime and k is an integer such that , then p divides

When I was much younger I found a proof of Fermat's Little Theorem using the following observation:

"Every -th number is divisible by ".

Now,

We want to show that,

Is an integer.

Now is not a -th number, nor a -rd number , ... , nor a -th number. Hence one of those numbers: and one of those is a -th thus,
is an integer.