Answer:
If [tex]n^2[/tex] is divisible by 3, the n is also divisible by 3.
Step-by-step explanation:
We will prove this with the help of contrapositive that is we prove that if n is not divisible by 3, then, [tex]n^2[/tex] is not divisible by 3.
Let n not be divisible by 3. Then [tex]\frac{n}{3}[/tex] can be written in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are co-prime to each other or in other words the fraction is in lowest form.
Now, squaring
[tex]\frac{n^2}{9} = \frac{x^2}{y^2}[/tex]
Thus,
[tex]n^2 = \frac{9x^2}{y^2}[/tex]
[tex]\frac{n^2}{3} = \frac{3x^2}{y^2}[/tex]
It can be clearly seen that the fraction [tex]\frac{3x^2}{y^2}[/tex] is in lowest form.
Hence, [tex]n^2[/tex] is not divisible by 3.
Thus, by contrapositivity if [tex]n^2[/tex] is divisible by 3, the n is also divisible by 3.
