Question

If a perfect square is divisible by prime p then it is also divisible by square of p

Answer 1

So if a perfect square is divisible by a prime p then it is also divisible by the square of p. This is the conversion of this sentence into the predicate form. ... Since that is every perfect square so for all x perfect square x there is some prime number which divides it.

Answer 2

Answer:

If $a$ perfect square is divisible by prime $p$ then it is also divisible by square of $p$.

Step-by-step explanation:

Given: If a perfect square is divisible by prime p then it is also divisible by square of p.

Give reason.

This statement is true.

Let $a$ be perfect square then $a=b^{2}$ for some $b$

Also, $a$ is divisible by prime $p$ then we can say that $p$ is factor of $a$ this implies $p$ is also factor of $b$.

So, $b^{2} =(pq)^{2}$, $q$ is some other factor.

$b^{2} =p^{2}q^{2}$ clearly it is divisible by $p^{2}$ which gives $a$ is divisible by $p^{2}$.

Hence, if $a$ perfect square is divisible by prime $p$ then it is also divisible by square of $p$.

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