Question

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

Answer 1

From euclids divison algorithm

a=bq+r-1

Where r is greater than equal to 0 and less than b

Substituting b=4 in equation 1

r=0,1,2,3

When r=0,a=4m

When r=1,a=4m+1

Whenr=2,a=4m+2

Whenr=3,a=4m+3

Any positive integer can be of the form 4m,4m+1,4m+2,4m+3

But 4m and 4m+2 are even

Hence any positive integer is of the 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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