Question

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

Answer 1

let 'a' be any positive integer then
a=bq+r; 0<equal to 4< r
for 'r' =0
a=4p
a=2(2p)
a=2m;which is even positive intgr...
for 'r'=1
a=4p+1
a=2(2p)+1
a=2m+1;which is odd positive intgr...
for'r'=2
a=4p+2
a=2(2p+1)
a=2m;which is a even positive intgr...
for'r'=3
a=4p+3
a=4p+2+1
a=2(2p+1)+1
a=2m+1;which is odd positive intgr...
HENCE,ANY ODD POSITIVE ITGR... CAN BE WRITTEN IN FORM (4m+1) OR (4m+3).

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

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