Question

plz solve it with full steps.. no short cut..
solve it..

quality answers needed..

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Answer 1

${\sf{\underline{\underline{\pink{Solution:-}}}}}$

If p² is an even integer, then p is an even.

${\sf{\underline{\underline{\pink{Reason:-}}}}}$

As we know square of even no. is always even.

Answer 2

$\bf{\underline{Question:-}}$

• If p$\sf ^2$ is an integer , then P is

$\bf{\underline{Given:-}}$

• p$\sf ^2$ is an integer

$\bf{\underline{To\:Find:-}}$

• P is an = ?

$\bf{\underline{Solution:-}}$

Let us assume,

→ $\sf P:n^2 \: is\: an\:even\: integer$

→ $\sf q : n \: is\: an\:even\: integer$

→ P $\sf ^2$ is true ( statement )

• n is not even integer
• n$\sf ^2$ is also not be even integer

So,

• Our contradiction is wrong
• P is not true

$\bf{\underline{Therefore:-}}$

• if P is even then q is also even

__________________________

$\sf p^2 \: is \:even \:integer$

BeCause:-

Square of even number is always even

Example

• $\bf 2^2 = 4$
• $\bf 4^2 = 16$
• $\bf 6^2 = 36$