Math, asked by mahekkhan58, 8 months ago

show that any positive odd integers is of the form 6q+1 or 6q+3 or 6q+5 where q is some integers ​

Answers

Answered by MysteriousAryan
3

Answer:

Let 'a' be any positive integer and b=6

Apply Euclid division lemma to A and B

a = 6q + r \: \:  \:  \:  where \: 0   \leqslant r < 6

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

a=6q,6q+1,6q+2,6q+3,6q+4,6q+5

and. a is positive odd integer

a≠6q. or a≠ 6q+2 or a≠6q+4

And. a=6q+1 ,a=6q+3 , a=6q+5

Hence proved

Answered by llTheUnkownStarll
0

 \huge \fbox \red{Solution:}

Let ‘a’ be any positive integer.

Then from Euclid's division lemma,

a = bq+r ; where 0 < r < b

Putting b=6 we get,

⇒ a = 6q + r, 0 < r < 6

For r = 0, we get a = 6q = 2(3q) = 2m, which is an even number. [m = 3q]

For r = 1,

We get a = 6q + 1 = 2(3q) + 1 = 2m + 1, which is an odd number. [m = 3q]

For r = 2,

We get a = 6q + 2 = 2(3q + 1) = 2m, which is an even number. [m = 3q + 1]

For r = 3,

We get a = 6q + 3 = 2(3q + 1) + 1 = 2m + 1, which is an odd number. [m = 3q + 1]

For r = 4,

We get a = 6q + 4 = 2(3q + 2) + 1 = 2m + 1, which is an even number. [m = 3q + 2]

For r = 5,

We get a = 6q + 5 = 2(3q + 2) + 1 = 2m + 1, which is an odd number. [m = 3q + 2]

Thus, from the above it can be seen that any positive odd integer can be of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.

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