Question

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.​

Answer 1

Answer:

Using Euclid division algorithm, we know that a=bq+r, 0≤r≤b ----(1)

Let a be any positive integer and b=6.

Then, by Euclid’s algorithm, a=6q+r for some integer q≥0, and r=0,1,2,3,4,5 ,or 0≤r<6.

Therefore, a=6qor6q+1or6q+2or6q+3or6q+4or6q+5

6q+0:6 is divisible by 2, so it is an even number.

6q+1:6 is divisible by 2, but 1 is not divisible by 2 so it is an odd number.

6q+2:6 is divisible by 2, and 2 is divisible by 2 so it is an even number.

6q+3:6 is divisible by 2, but 3 is not divisible by 2 so it is an odd number.

6q+4:6 is divisible by 2, and 4 is divisible by 2 so it is an even number.

6q+5:6 is divisible by 2, but 5 is not divisible by 2 so it is an odd number.

And therefore, any odd integer can be expressed in the form 6q + 1 or 6q + 3 or 6q + 5

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Answer 2

Answer:

Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5; where q is some integer. Answer: According to Euclid's Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r < b.