Question

show that any positive odd integer is of the form (4m+1) or(4m+3) where m is some integer

Answer 1

Hii friend,

Let n be an arbitrary positive integer.

On dividing n by 4 , let m be the Quotient and r be the remainder.

By Euclid's division lemma, we have


n = 4m +r. where r = 0,1,2,3

Therefore,

n = 4m or (4m+1) or (4m+2) or (4m+3)

Clearly , 4m and (4m+2) are Even and since n is odd n will not equal to 4m and 4m+2

Therefore,

n = (4m+1) or (4m+3) , for some integer m.

Hence,

any positive odd integer is in the form of (4m+1) or (4m+3) for some integer m.

HOPE IT WILL HELP YOU.... :-)

Answer 2

Step-by-step explanation:

Note :- I am taking q as some integer.

Let a be the positive integer.

And, b = 4 .

Then by Euclid's division lemma,

We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .

Taking r = 0 .

a = 4q .

Taking r = 1 .

a = 4q + 1 .

Taking r = 2

a = 4q + 2 .

Taking r = 3 .

a = 4q + 3 .

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved

THANKS

#BeBrainly.