Question

use euclid's Lemma to show that any positive odd integer is of the form 4 Q + 1, 4Q + 3

Answer 1

a = bq + r
b = 4
r = 0,1,2,3

I ) When r = 0

a = 4q + 0

a = 4q ( Which is not in the form of 4q + 1 or 4q+3 )

II ) When r = 1

a = 4q + 1 ( Which is in the form of 4q +1 )

III ) When r = 2

a = 4q + 2 (Which is not in the form of 4q + 1 or 4q+3)

IV ) When r = 3

a = 4q + 3 ( Which is in the form of 4q+3 )

V ) When r = 4

a = 4q + 4 ( Which is not in the form of 4q + 1 or 4q+3 )


Therefore, we can conclude that any positive odd integer is in the form of 4q + 1 or 4q + 3



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Answer 2

Step-by-step explanation:

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 .