Question

The square of an odd number is
A. always an even number
B. sometimes even and sometimes odd
C. always an odd number
D. always an irrational number

Answer 1

Hint: This question can be solved by assuming that the odd number is always of the form 2n+1 where n is even and the even number is of the form 2n where n can be even or odd. By assuming this and continuing i.e. squaring the odd form of a number we can get to the answer.

Complete step-by-step answer:
It is known that the odd number is always of the form 2n+1
Let $I = 2n + 1$ where n is even
On squaring both sides,
$ \Rightarrow {I^2} = {\left( {2n + 1} \right)^2}$
On expanding further,
$ \Rightarrow {I^2} = 4{n^2} + 2n + 1$
Taking 2n common from RHS,
$ \Rightarrow {I^2} = 2n\left( {2n + 1} \right) + 1$
$\because {\text{ }}I = 2n + 1$
On substituting,
$ \Rightarrow {I^2} = 2nI + 1$
$\because $ I is odd and on multiplying it with 2 it becomes even,
$\Rightarrow$ 2nI is even and 2nI+1 is odd
$\Rightarrow$ ${I^2}$ is odd
After squaring for a finite number of times it can be drawn that the result is always an odd number.
$\Rightarrow$ On squaring an odd number the result is always an odd number.
So, the correct answer is “Option C”.

Note: In this question the general form any number must be remembered to solve the question. In this general form the odd number taken is 2n+1 but it can also be taken as 2n-1 the result would be the same. Always try to solve the question step by step so that the wrong step can be figured out quickly. If the calculations are wrong then the final result would not come.