Question

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

Answer 1

Let Q be odd integer..
On dividing Q by 4 , Let m be the quotient and r be the remainder.
So,by Euclid's division lemma , We have
Q =4m + r Where r = 1,2,3
Therefore,
Q = 4m or (4m+1) or (4m+2) or (4m+3) .
Clearly,
4m and (4m+2) are even and since Q is odd ,so Q is not equal to 4m and Q not equal to 4m+2.
Therefore,
Q = 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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