Math, asked by Khushi1511, 1 year ago

Show that square of any positive odd integer is of the form 8q+1,where q is any positive integer.

Answers

Answered by GovindRavi
2
hope you get the solution....if doubt arises in any steps..txt me in comments..
Attachments:

Khushi1511: Use Euclid's division lemma to show that the cube of any positive integer is of the form 9q,9q+1,9q+8, where q is some integer.
GovindRavi: i cant type the whole solution for the same problem...evwn there is no uploading option here
Khushi1511: ohh
GovindRavi: just upload the same prob on brainly publicly
Khushi1511: ok
GovindRavi: once you done txt me
Khushi1511: yeah done
GovindRavi: kk wait
Answered by Anonymous
6

Step-by-step explanation:


Hey friends !!




Let 'a' be the any positive integer.


Then, b = 8 .


Using Euclid's division lemma :-


0 ≤ r < b => 0 ≤ r < 8 .


•°• The possible values of r is 0, 1, 2, 3, 4, 5, 6, 7.


▶ Question said Square of odd positive integer , then r = 1, 3, 5, 7 .



→ Taking r = 1 .


a = bm + r .


= (8q + 1)² .


= 64m² + 16m + 1


= 8( 8m²+ 2m ) + 1 .


= 8q + 1 . [ Where q = 8m² + 2m ]


→ Taking r = 3 .


a = bq + r .


= (8q + 3)² .


= 64m² + 48m + 9 = 64m² + 48m + 8 + 1 .


= 8( 8m²+ 6m + 1 ) + 1 .


= 8q + 1 . [ Where q = 8m² + 6m + 1 ]



→ Taking r = 5 .


a = (8q + 5)² .


= 64m² + 80m + 25 = 64m² + 80m + 24 + 1 .


= 8( 8m²+ 10m + 3 ) + 1 .


= 8q + 1 . [ Where q = 8m² + 10m + 3 ]



→ Taking r = 7 .


a = ( 8q + 7 )² .


= 64m² + 112m + 49 = 64m² + 112m + 48 + 1 .


= 8( 8m²+ 14m + 6 ) + 1 .


= 8q + 1 . [ Where q = 8m² + 14m + 6 ] .



Hence, the square of any odd positive integer is of the form 8q + 1 .


✓✓ Proved ✓✓




THANKS



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