Show that the square of an odd positive integer is of the form 8m + 1, for some whole number m
Solution:
Using the Euclid division lemma , a = bq + r where 0 ≤ r < b
Consider b = 8 and r ∈ [1, 7] means r = 1, 2, 3, .....7
Here, a = 8q + r
Case 1 - If r = 1 , a = 8q + 1
By squaring on both sides,
a2 = (8q + 1)2
= 642q2 + 16q + 1
= 8(8q2 + 2q) + 1
= 8m + 1 , where m = 8q2 + 2q
Case 2 - If r = 2 , a = 8q + 2
By squaring on both sides,
a2 = (8q + 2)2
= 64q2 + 32q + 4 ≠ 8m +1 [When r is an even number it is not of the form of 8m + 1]
Case 3 - If r = 3 , a = 8q + 3
By squaring on both sides,
a2 = (8q + 3)2
= 64q2 + 48q + 9
= 8(8q2 + 6q + 1) + 1
= 8m + 1 , where m = 8q2 + 6q + 1
Here every odd value of r square of a is of the form 8m +1
At every even value of r, the square of a is not in the form of 8m + 1 .
We know that,
a = 8q + 1 , 8q + 3 , 8q + 5 , 9q + 7 are not divisible by 2
It means that all numbers are odd numbers
Therefore, the square of an odd positive is of the form 8m + 1, for some whole number m.
✦ Try This: The sum of 3 consecutive odd numbers is 21. Find the 3 numbers using the odd numbers formula
Let the 3 consecutive odd numbers be 2n + 1, 2n + 3 and 2n + 5.
Given: (2n + 1) + (2n + 3) + (2n + 5) = 21
6n + 9 = 21
6n = 12
n = 2
Thus 2n + 1 = 5
2n + 3 = 7
2n + 5 = 9
Therefore, the 3 consecutive odd numbers are 5, 7 and 9
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 1
NCERT Exemplar Class 10 Maths Exercise 1.3 Sample Problem 2
Show that the square of an odd positive integer is of the form 8m + 1, for some whole number m
Summary:
The square of an odd positive integer is of the form 8m + 1, for some whole number m
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