Question

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

Answer 1

Answer:

let a be any positive integer and b=4

by using Euclid division lemma

a = bq + r

where b > r >= 0

so values of r will be 0,1,2 and 3

hence we can write

a = 4q

a=4q+1

a=4q+2

a=4q+3

we know that 4m and 4m+2 is even no. because it has common factor 2.

So, we find that 4m+1 and 4m+3 is odd no. because is no. is either even or odd.

I hope it will help u

Answer 2

Answer:

Step-by-step explanation:

Let, a be any odd positive integer and b =4 . By division lemma there exists integers m and n such that,

a=4m+n,where 0 <n <4

a=4m or a=4m+1 or a=4m+2 or a= 4m +3

(Because 0 < n < 4)

a=4m+1 or 4m+3

(Because a is an odd integer )

Hence 4m+1 or 4m+3 are any odd integers