Question

Show that any positive integer is the form 4qor 4q+2 where q is a whole number ​

Answer 1

Answer:

Any positive integer can be represented as 

4q+a

  where q is a integer

              a can be 0,1,2,3

if a=0

    then number is 4q 

          so it is divisible by 2

                    So it is an even number

 if a=2

         then number is 

4q+2=2(2q+1)4q+2=2(2q+1)

              so it is divisible by 2

                    So it is an even number

    Hence proved

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Answer 2

Answer:

Let positive integer a = 4m + r , By division algorithm we know here 0 ≤ r < 4 , So

When r = 0

a = 4m

Squaring both side , we get

a2 = ( 4m )2

a2 = 4 ( 4m2)

a2 = 4 q , where q = 4m2

When r = 1

a = 4m + 1

squaring both side , we get

a2 = ( 4m + 1)2

a2 = 16m2 + 1 + 8m

a2 = 4 ( 4m2 + 2m ) + 1

a2 = 4q + 1 , where q = 4m2 + 2m

When r = 2

a = 4m + 2

Squaring both hand side , we get

a2 = ( 4m + 2 )2

a2 = 16m2 + 4 + 16m

a2 = 4 ( 4m2 + 4m + 1 )

a2 = 4q , Where q = 4m2 + 4m + 1

When r = 3

a = 4m + 3

Squaring both hand side , we get

a2 = ( 4m + 3)2

a2 = 16m2 + 9 + 24m

a2 = 16m2 + 24m + 8 + 1

a2 = 4 ( 4m2 + 6m + 2) + 1

a2 = 4q + 1 , where q = 4m2 + 6m + 2

Hence

Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.