Q. 24.0( 854 Votes )

Show that any pos

Answer :

To Prove: Any Positive odd integer is of the form 6q + 1, 6q + 3, 6q + 5
 
Proof: To prove the statement by Euclid's lemma we have to consider divisor as 6 and then find out the possible remainders when divided by 6
 
By taking,’ a’ as any positive integer and b = 6.

Applying Euclid’s algorithm

a = 6 q + r
As divisor is 6 the remainder can take only 6 values from 0 to 5

Here,  r = remainder = 0, 1, 2, 3, 4, 5 and q ≥ 0

So, total possible forms are  6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5

6q + 0 , (6 is divisible by 2, its an even number)

6q + 1, ( 6 is divisible by 2 but 1 is not divisible by 2, its an odd number)

6q + 2, (6 and 2 both are divisible by 2, its an even number)

6q + 3, (6 is divisible by 2 but 3 is not divisible by 2, its an odd number)

6q + 4, ( 6 and 4 both are divisible by 2, its an even number)

6q + 5 , (6 is divisible by 2 but 5 is not divisible by 2, its an odd number)

Therefore, odd numbers will be in the form 6q + 1, or 6q + 3, or 6q+5
Hence, Proved.

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