Question

Use the method of direct proof to prove the following statements. If \(p\) is prime and \(k\) is an integer for which \(0

Short Answer

Expert verified

Given that \(p\) is prime and \(k\) is an integer for which \(0<k<p\), it can be concluded through the method of direct proof that \(p\) indeed divides \(\left(\begin{array}{l}p \ k\end{array}\right)\).

Step by step solution

Definition of the binomial coefficient

The binomial coefficient \(\left(\begin{array}{l}p \ k\end{array}\right)\), also written as 'p choose k', is defined as \(\left(\begin{array}{l}p \ k\end{array}\right)=\frac{p!}{k!(p-k)!}\) where \(p!\) denotes ‘p-factorial’, the product of all positive integers less than or equal to \(p\).

Rewrite the binomial coefficient

To check whether \(p\) divides \(\left(\begin{array}{l}p \ k\end{array}\right)\), the binomial coefficient \(\left(\begin{array}{l}p \ k\end{array}\right)\) can be rewritten as \(\frac{p*(p-1)*...*(p-k+1)}{k*(k-1)*...*1}\). Now we have \(p\) as part of a number in the numerator. Next, need to show that \(p\) divides this number exactly (no remainder).

Check divisibility by \(p\)

Because \(p\) is a prime number, it only has 1 and itself as divisors. For all \(k\) (where \(0<k<p\)), it means \(k\) cannot divide \(p\) and moreover, since the definition of factorial is a product of all positive integers less than or equal to \(k\), no number in the denominator will divide \(p\). Therefore, the prime number \(p\) will not divide the denominator but will divide the numerator. Therefore, \(p\) does indeed divide \(\left(\begin{array}{l}p \ k\end{array}\right)\). This proves our conjecture.

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Short answer

Given that \(p\) is prime and \(k\) is an integer for which \(0<k<p\), it can be concluded through the method of direct proof that \(p\) indeed divides \(\left(\begin{array}{l}p \ k\end{array}\right)\).

Step-by-step solution

Definition of the binomial coefficient

The binomial coefficient \(\left(\begin{array}{l}p \ k\end{array}\right)\), also written as 'p choose k', is defined as \(\left(\begin{array}{l}p \ k\end{array}\right)=\frac{p!}{k!(p-k)!}\) where \(p!\) denotes ‘p-factorial’, the product of all positive integers less than or equal to \(p\).

Rewrite the binomial coefficient

To check whether \(p\) divides \(\left(\begin{array}{l}p \ k\end{array}\right)\), the binomial coefficient \(\left(\begin{array}{l}p \ k\end{array}\right)\) can be rewritten as \(\frac{p*(p-1)*...*(p-k+1)}{k*(k-1)*...*1}\). Now we have \(p\) as part of a number in the numerator. Next, need to show that \(p\) divides this number exactly (no remainder).

Check divisibility by \(p\)

Because \(p\) is a prime number, it only has 1 and itself as divisors. For all \(k\) (where \(0<k<p\)), it means \(k\) cannot divide \(p\) and moreover, since the definition of factorial is a product of all positive integers less than or equal to \(k\), no number in the denominator will divide \(p\). Therefore, the prime number \(p\) will not divide the denominator but will divide the numerator. Therefore, \(p\) does indeed divide \(\left(\begin{array}{l}p \ k\end{array}\right)\). This proves our conjecture.