Question

show that any +ve integer is in the former of 8q+1 , 8q+3 ,8q+5, 8q+7 where quality is some integer

Answer 1

let a be any positive integer and b =8 ,then , ecluid algorithm ,

a = 8q +r for some integer q >or = 0 and r = 0,1,2,3,4,5,6,7 0 <,= r <= 7

then , 8q +1 = 2 x 4q +1 => 2m +1, where m is any positive integer = 4q

8q+3 = 8q +2+1 => 2 (4q +1) +1 => 2 m +1 , where m is any positive integer = 4q +1

8q +5 = 8q + 4+1 => 2 (4q +2) +1 => 2 m +1 , where m is any positive integer = 4q+2

8q +7 = 8q + 6+1 => 2( 4q +3) +1 => 2m +1 , where m is any positive integer = 4q +3

this means that 8q +1, 8q+3 , 8q+5 and 8q+7 cannot be exactly divisible by 2 . hence any odd no. can be expressed in the form of 8q+1 , 8q+3 , 8q+5 and 8q+7


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Answer 2

Answer:

Yes

Let a be any positive integer

According to Euclids division lemma

a = bq + r

b = 8

so the possible values of r = 0,1,2,3,4,5,6,7

a = 8q ( it is even)

a = 8q + 1 ( it is odd)

..........and so on

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