To prove that the square of an odd number is an odd number, let's consider an odd number in a general form. An odd number can be expressed as [tex]\(2n + 1\)[/tex], where [tex]\(n\)[/tex] is an integer.
We need to examine the square of this odd number:
1. Express the odd number and then square it:
[tex]\[
(2n + 1)^2
\][/tex]
2. Apply the binomial theorem to expand the expression:
[tex]\[
(2n + 1)^2 = (2n + 1)(2n + 1)
\][/tex]
3. Use the distributive property to expand the multiplication:
[tex]\[
(2n + 1)(2n + 1) = 2n(2n) + 2n(1) + 1(2n) + 1(1)
\][/tex]
4. Simplify each term:
[tex]\[
= 4n^2 + 2n + 2n + 1
\][/tex]
5. Combine like terms to get the final expanded expression:
[tex]\[
= 4n^2 + 4n + 1
\][/tex]
So, we have:
[tex]\[
(2n + 1)^2 = 4n^2 + 4n + 1
\][/tex]
Notice that [tex]\(4n^2 + 4n\)[/tex] is always even because it is a multiple of 4, and adding 1 to an even number results in an odd number. Therefore, the final expression [tex]\(4n^2 + 4n + 1\)[/tex] is always odd.
Hence, squaring any odd number results in an odd number.✅
