Question

Question: Let pbe a prime and $\mathit{k}\mathbf{,}\mathit{a}\mathbf{\in }\mathbf{\mathbb{Z}}$such that$\mathit{p}\mathbf{\nsim }\mathit{a}$ and 0 < k < p . Prove that$\mathbf{ka}\not\equiv \mathbf{0}\left(\mathbf{modp}\right)$ .[Hint: Theorem 1.5]

Short answer

Answer

It can be concluded that $ka\not\equiv 0\left(\mathrm{mod}p\right)$ is proved using contradiction.

Step-by-step solution

Define the property satisfying an integer  p to be a prime.

If an integer p satisfies the following property, then p is said to be a prime.

Whenever p divide st , then it must divide either s or t …(1)

Suppose integers p and q are prime numbers. If one of them divides the other, then the equality $p=\pm q$ must be true. …(2)

Prove the required result by the method of proof of contradiction

Assume that the following equation istrue:

$ka\not\equiv 0\left(\mathrm{mod}p\right)$

This congruence equation implies that p must divide the product ka . So, according to the statement (1), the prime number p must divide either k or a.

Considering that the prime number p divide k .

Now, noting the inequality.

This implies that the integer k is smaller than the prime number.

Thus, a bigger number p cannot divide a smaller numbercompletely, because the result of the division must be a fractional number.

This implies that p must divide.

Which is a contradiction to the given fact that p does not divide a .

Therefore, our assumption is incorrect.

Hence, the required result is proved.