Can this be considered a proof

  • Thread starter Government$
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In summary, the conversation discusses the divisibility of p by the combination of p and k, given that p is a prime and k is an integer between 0 and p. The conclusion is that p will always divide the combination of p and k, as shown by the proof using Gauss' theorem.
  • #1

Homework Statement


If p is a prime and k is an integer for which 0<k<p, then p divides [itex]\displaystyle \ \binom{p}{k}[/itex].
Whne p divides [itex]\displaystyle \ \binom{p}{k}[/itex] it means that [itex]\displaystyle \ \binom{p}{k}=p*b[/itex].
wheren b is some number.

The Attempt at a Solution



So p is equal to some number k.

[itex]p=k[/itex]
I take factorial of both sides and get
[itex]p!=k![/itex]
and multiply both sides by
[itex]\frac{1}{k!(p-k)!}[/itex]
to get
[itex]\frac{p!}{k!(p-k)!}=\frac{k!}{k!(p-k)!}[/itex]
Since p=k
[itex]\displaystyle \ \binom{p}{k}=\frac{p(p-1)(p-2)...(p-k+1)(p-k)!}{k!(p-k)!}[/itex]
then
[itex]\displaystyle \ \binom{p}{k}=p*\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex]
where [itex]b=\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex] therefore p divides [itex]\displaystyle \ \binom{p}{k}[/itex].
--------------------------------------------------------------------------------------

Homework Statement



If [itex]n \in N[/itex] then [itex]\displaystyle \ \binom{2n}{n}[/itex] is even i.e.
[itex]\displaystyle \ \binom{2n}{n}=2*b[/itex] where b is some number

The Attempt at a Solution



[itex]\displaystyle \ \binom{2n}{n}=\frac{2n!}{n!(2n-n)!} = \frac{2n(2n-1)..3*2*1}{n!(2n-n)!}=2*\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex] where [itex]b=\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex]
 
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  • #2
Government$ said:

Homework Statement


If p is a prime and k is an integer for which 0<k<p, then p divides [itex]\displaystyle \ \binom{p}{k}[/itex].
Whne p divides [itex]\displaystyle \ \binom{p}{k}[/itex] it means that [itex]\displaystyle \ \binom{p}{k}=p*b[/itex].
wheren b is some number.

The Attempt at a Solution



So p is equal to some number k.

[itex]p=k[/itex]
I take factorial of both sides and get
[itex]p!=k![/itex]
and multiply both sides by
[itex]\frac{1}{k!(p-k)!}[/itex]
to get
[itex]\frac{p!}{k!(p-k)!}=\frac{k!}{k!(p-k)!}[/itex]
Since p=k
[itex]\displaystyle \ \binom{p}{k}=\frac{p(p-1)(p-2)...(p-k+1)(p-k)!}{k!(p-k)!}[/itex]
then
[itex]\displaystyle \ \binom{p}{k}=p*\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex]
where [itex]b=\frac{(p-1)(p-2)...(p-k+1)}{k!}[/itex] therefore p divides [itex]\displaystyle \ \binom{p}{k}[/itex].
--------------------------------------------------------------------------------------

Homework Statement



If [itex]n \in N[/itex] then [itex]\displaystyle \ \binom{2n}{n}[/itex] is even i.e.
[itex]\displaystyle \ \binom{2n}{n}=2*b[/itex] where b is some number

The Attempt at a Solution



[itex]\displaystyle \ \binom{2n}{n}=\frac{2n!}{n!(2n-n)!} = \frac{2n(2n-1)..3*2*1}{n!(2n-n)!}=2*\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex] where [itex]b=\frac{n(2n-1)..3*2*1}{n!(2n-n)!}[/itex]

In 1): how do you know that
[tex]b=\frac{(p-1)(p-2)...(p-k+1)}{k!} [/tex]
is an integer? Where have you used the hypothesis that p is prime?
 
  • #3
Ray Vickson said:
In 1): how do you know that
[tex]b=\frac{(p-1)(p-2)...(p-k+1)}{k!} [/tex]
is an integer? Where have you used the hypothesis that p is prime?

Good points, there goes my proof in the water.:cry:
 
  • #4
At the start of 1 you write p=k even though p > k...

Your proof isn't totally wrecked, you know that
[tex] \frac{ p(p-1)(p-2)...(p-k+1)}{k!} [/tex]
is an integer, so all you want to do is explain why, when you divide the numerator by k!, the p in the numerator is not canceled out.
 
  • #5
Government$ said:
Good points, there goes my proof in the water.:cry:

Just to clarify: it is important that p be prime. For example (writing C(a,b) for "a choose b") we have that C(5,2) = 10 is divisible by 5 and C(7,3) = 35 is divisible by 7, but neither C(6,2) = 15 nor C(6,3) = 20 are divisible by 6.
 
  • #6
The hypotheses say 0< k< p yet in your "proof" you say "p is equal to some number k". Surely p is equal to "some number" (why not just use "p"?) but calling it "k" and then confusing it with the "k" in the hypothesis is just "sleight of hand". I saw you pocket that card!
 
  • #7
Hi,

It's a consequence of Gauss theorem.
Write ## p! = k!\ (p-k)!\ C_p^k##. You get that ##p|k!\ (p-k)!\ C_p^k##.
Show that ##\text{gcd}(p, k!\ (p-k)!) = 1 ##, and conclude ##p|C_p^k##
 

1. Can this be considered a proof?

It depends on the context and the specific criteria for what constitutes a proof. In science, a proof is typically a logical and empirical demonstration of a hypothesis or theory. Therefore, the validity of a proof depends on its ability to support or refute a specific scientific claim.

2. What is required for something to be considered a proof?

A proof must be based on sound logic and empirical evidence. It should also be replicable, meaning that it can be repeated by other scientists and yield the same results. Additionally, a proof should be able to withstand scrutiny and be supported by multiple lines of evidence.

3. How can I determine if a proof is valid?

To determine the validity of a proof, you should evaluate the methodology and evidence used to support the claim. Look for any potential flaws or biases in the study, and consider if the results are consistent with other research in the field. Peer review is also an important step in validating a proof.

4. Is a proof always definitive?

No, a proof is not always definitive and is subject to change as new evidence and knowledge is discovered. In science, theories and hypotheses are constantly refined and updated as our understanding of the natural world evolves. Therefore, a proof should be considered as the most current and well-supported explanation at a given time.

5. Can a proof ever be considered absolute?

In science, nothing can be considered absolute or certain. This is because new evidence and knowledge can always challenge or modify existing theories and explanations. However, a proof can be considered the best explanation based on the available evidence and current understanding of the natural world.

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