Question

2. Using Euclid's Lemma show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

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Answer 1

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Let n be any positive integer

let b = 4(divisor)

By Euclid's Division Lemma

=> n = 4q + r where 0<=r<b

Thus r = 0,1,2 or 3

Case 1, when r = 0

=> n = 4q

=> n = 2(2q) (even)(because divisible by 2)

Case 2, when r = 1

=> n = 4q + 1 (odd)(not divisible by 2)

Case 3, when r = 2

=> n = 4q + 2

=> n = 2(2q + 1)

=> n = 2m where m = 2q + 1 << (even) (divisible by 2)

Case 4, when r = 3

=> n = 4q + 3 (odd) (not divisible by 2)

Here from Case 2 and Case 4

We can say that any positive odd integer is of the form 4q + 1 or 4q + 3

Hence Proved

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Hope this helps ✌️

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 .