Show that the square of an odd +ve integer is of the form 8m+1.
Answers
Answer:
Answer:Squares of odd numbers are odd, since (2n + 1)2 = 4(n2 + n) + 1. It follows that square roots of even square numbers are even, and square roots of odd square numbers are odd. As all even square numbers are divisible by 4, the even numbers of the form 4n + 2 are not square numbers.
Answer:Squares of odd numbers are odd, since (2n + 1)2 = 4(n2 + n) + 1. It follows that square roots of even square numbers are even, and square roots of odd square numbers are odd. As all even square numbers are divisible by 4, the even numbers of the form 4n + 2 are not square numbers.Explanation:
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Answer:
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Explanation:
According to Euclid division lemma , a = bq + r where 0 ≤ r < b
Here we assume b = 8 and r ∈ [1, 7 ] means r = 1, 2, 3, .....7
Then, a = 8q + r
Case 1 :- when r = 1 , a = 8q + 1
squaring both sides,
a² = (8q + 1)² = 64²q² + 16q + 1 = 8(8q² + 2q) + 1
= 8m + 1 , where m = 8q² + 2q
case 2 :- when r = 2 , a = 8q + 2
squaring both sides,
a² = (8q + 2)² = 64q² + 32q + 4 ≠ 8m +1 [ means when r is an even number it is not in the form of 8m + 1 ]
Case 3 :- when r = 3 , a = 8q + 3
squaring both sides,
a² = (8q + 3)² = 64q² + 48q + 9 = 8(8q² + 6q + 1) + 1
= 8m + 1 , where m = 8q² + 6q + 1
You can see that at every odd values of r square of a is in the form of 8m +1
But at every even Values of r square of a isn't in the form of 8m +1 .
Also we know, a = 8q +1 , 8q +3 , 8q + 5 , 9q +7 are not divisible by 2 means these all numbers are odd numbers
Hence , it is clear that square of an odd positive is in form of 8m +1