Question

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

Answer 1

Answer:

Step-by-step explanation:

by euclid's division lemma

a=bq+r

b=6

therefore r=0,1,2,3,4,5

r=0

a=bq+r

a=6q+0=6q=even

r=1

a=6q+1=odd -(1)

r=2

a=6q+2=even

r=3

a=6q+3=odd -(2)

r=4

a=6q+4=even

r=5

a=6q+5=odd -(3)

by 1,2 and 3 we get

6q+1,6q+3,6q+5=negative positive integer where q is some integer

hence proved

Answer 2

$\huge \underline \mathbb {SOLUTION:-}$

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, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.