Prove that the square of an odd integer is always odd
This article focuses on discussing in detail the proof of why the square of an odd integer is always an odd number.
A number is said to be odd if it is not divisible by 2, or if a number can be expressed in the form of (2k +1), for some integer k, then the number is called an odd number.
Given two numbers A and B, if A * A = B then, A is known as the square root of B.
The square of an odd integer is always an odd number.
This section discusses the proof of the above problem statement-
1. Consider an odd integer, X. According to the above definition, A can be written as-
X = (2k + 1), for some integer k
2. Now, squaring both sides-
X2 = (2k + 1)2 ---(1)
3. The formula for the square of the sum of 2 numbers is-
(A + B)2 = A2 + 2AB + B2
4. Using the above property in equation (1)-
X2 = (2k)2 + 4k + 1 X2 = 4k2 + 4k + 1 ---(2)
5. Now, let’s do some rearrangements in equation 2 as-
X2 = 2(2k2 + 2k) + 1
6. Notice the right-hand side of the above equation. Since K is an integer, (2k2 + 2k) is also an integer. Now, let’s assume an integer, m = (2k2 + 2k). The above equation can be written as-
X2 = (2m + 1), for some integer m
7. From the above equation and the definition of an odd integer, it can be concluded that X2 is also an odd integer, which proves our statement that the Square of an odd integer is always odd.
For X = 3-
1. Put the value X = 3 in the above equations’ step by step-
X = (2k + 1), for some integer k 3 = (2k + 1), for k = 1 (integer)
2. Now, if square of X is taken-
X2 = (2k + 1)2 X2 = 4K2 + 4K + 1
3. Again, after doing arrangements, X2 can be written as-
X2 = 2(2k2 + 2k) + 1 for X = 3, 9 = 2(2k2 + 2k) + 1
4. For k = 1, (2k2 + 2k) evaluates to 4. Let m= (2k2 + 2k) = 4 i.e.
9 = 2m + 1, for m = 4 (integer)
Now, from the above definition of an odd integer, it can be said that 9 is an odd number, which implies the square of an odd integer (In this case, 3) is always odd. Hence, Proved.
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