Question

show that any positive odd integer is of the form 8q+1 ,8q+3,8q+5 and 8q+7

Answer 1

Answer:

Step-by-step explanation:

let us start with with taking a, where a is a positive odd integer.

We apply the division algorithm with a and b = 8.

since 0 ≤ r < 8 the possible remainders are 0,1,2,....7.

That is a can be 8q or8q+1 or 8q+2 or 8q+3 or 8q+4or 8q+5 or 8q+6 where q is quotient.

How ever since a is odd a cannt be 8q or 8q+2 or 8q+4 (since they are divisible by 2).

There fore, any odd integer is of the form 8q+1, 8q+3, 8q+5 or 8q+7.

or

Let n be a positive odd integer. We need to show that n can be written in any one of the form of 8q+1, 8q+3, 8q+5 or 8q+7

According to division algorithm,

we can write any number ‘a’ in the form

a = 8q + r

where q is any integer and 0 <= r <= 7. So r can be 0, 1, 2, 3, 4, 5, 6 or 7.

Thus, a can be written as

a = 8q

a = 8q+2

a = 8q+3

a = 8q+4

a = 8q+5

a = 8q+6

a = 8q+7

We need only odd numbers. Since 8q, 8q+2, 8q+4, and 8q+6 are divisible by 2, they are even numbers.

So any odd integer can be written as any one of the remaining forms which are (8q+1, 8q+3, 8q+5 or 8q+7.)

Answer 2

$\huge \bf \red{ \mid{\overline{ \underline{ANSWER}}} \mid}$

» let a be any positive integer

» then

» b=8

→ 0≤r<b = 0≤r<8

→ r=0,1,2,3,4,5,6,7

» case 1.

› r=0

a=bq+r

8q+0

8q

→ case 2.

» r=1

› a=bq+r

8q+1

→ case3.

» r=2

› a=bq+r

8q+2

→ case 4.

» r=3

› a=bq+r

8q+3

→ case 5.

» r=4

› a=bq+r

8q+4

→ case 6.

» r=5

› a=bq+r

8q+5

→ case7.

» r=6

› a=bq+r

8q+6

→ case 8

» r=7

› a=bq+r

8q+7