Question

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

Answer 1

Answer:

let a be any positive integer with b=6, we shall apply euclids division lemma

a=bq+r(0<=r<b)

possible values for r are0, 1,2,3,4 5.

put r=1.we get

a=bq+r

a=6q+1(so it's odd.)

now put r=3.we get

a=bq+r

a=6q+3(odd)

now let us take r=5.we get

a=bq+r

a=6q+5(odd)

hence proved that any positive odd integer is in the form of 6q+1, 6q+3or6q+5.

Answer 2

Let a be the any positive integer and b = 6.

then, By the Euclid's Division Lemma,

          a= 6q + r                  [∵ a = bq + r]

where, 6 ≤ 0 < r                  [  b ≤ 0 < r ]

then the possible remainders are 0, 1, 2, 3, 4 and 5.

So, we obtain following cases

i]   a= 6q

ii]  a= 6q + 1

iii] a= 6q + 2

iv] a= 6q + 3

v]  a= 6q + 4

vi] a= 6q + 5

in Case i] a= 6q            

=>            a = 2(3q)          

=> a is divisible by 2      

∴ It is an even integer.

In Case ii] a= 6q + 1

=> a is not divisible by 2

∴ a is an odd integer.

Case iii] a= 6q + 2

=>          a = 2(3q + 1)

=> a is divisible by 2

∴ a is an even integer.

Case iv] a= 6q + 3

=>  a is not divisible by 2

∴ a is an odd integer.

Case v] a= 6q + 4

=>          a= 2(3q + 2)

=>  a is divisible by 2

∴ It is an even integer.

Case vi] a= 6q + 5

=> a is not divisible by 2

∴ It is an odd integer.

From the above Cases, we conclude that

Any positive integer is of the form 6q + 1, 6q + 3 or 6q + 5.

____________________Hence Proved_________________________

Hope this helped you..