Question

show that any positive odd integer is of the form (6m+1) or (6m+3) or (6m+5), where m is some integer

Answer 1

Heya!


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

Use Euclid's algorithm , we have,

a = 6q + r

Where, r = remainder and value of q is either more or equal to 0, as r = 1 , 2 , 3 , 4, 5

This is because, 0 <= r < b and value of b = 6

Possible forms are,

6q + 0
6q + 1
6q + 2
6q + 3
6q + 4
6q + 5

6q + 0 , 6 is divisible by 2, so it's an even number of course.

6q + 1 , 1 isn't divisible by 2 as it's an odd number.

6q + 2 , 2 is divisible by 2.

6q + 3 , 3 isn't divisible by 2.

6q+4 , 4 is divisible by 2

6q+ 5 , 5 isn't divisible by 2

So, clearly, we saw that any positive odd integer is in form of 6m+1, 6m+3 , 6m+5.

Also , there is another method,but the above one is better.

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Hope it helps...!!!

Answer 2

SOLUTION:--

LET ,any positive integer
m= 1,2,3,4,5........................................infinite

given form , (6m+1),(6m+3),(6m+5)

if m=1 then
(6×1+1),(6×1+3),(6×1+5)
7, 9, 11, are odd integer

if m=2 then

(6×2+1),(6×2+3),(6×2+5)

13, 15, 17, are odd interger

if m=3 then
(6×3+1),(6×3+3),(6×3+5)
19 , 21, 23, are odd interger

similarly if m =infinity = 1/0

(6×1/0+1),(6×1/0+3),(6×1/0+5)

6/0+1 , 6/0+3 , 6/0 +5
infinty+1 , infinty+3 , infinty+5
here o is even and 6 even = even/even = even.....then here( infinity value ia even)
here 1 , 3, 5 are odd numbers
even+odd = odd then
[infinty+1 , infinty+3 , infinty+5] are odd integer.


that proved :- any positive odd integer is of the form (6m+1) or (6m+3) or (6m+5), where m is some integer