Math, asked by harshitshrivastava59, 6 months ago

Show that th square of an odd positive integer is of the form 8m+1 ,where m is some whole number​

Answers

Answered by mohit810275133
2

Step-by-step explanation:

HEY MATE ..

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 +1But 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

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