Question

Using Euclid's division lemma for any
natural no. and 4 which of the
forms can a positive integer be
represented as where m is an integer?
a) 4m b)4m+ 1 c) 4m+2 d) All of the above​

Answer 1

Answer:

Hi friend,

1.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 . ( Hence proved )

2.Let positive integer a = 5m+ r , By division algorithm we know here 0 ≤ r < 5 , So

When r = 0

a = 5m

Squaring both side , we get

a2 = ( 5m )2

a2 = 5 ( 5m2 )

a2 = 5 q , where q = 5m2

When r = 1

a = 5m + 1

squaring both side , we get

a2 = ( 5m + 1 )2

a2 = 25m2 + 1 + 10m

a2 = 5 ( 5m2 + 2m ) + 1

a2 = 5q + 1 , where q = 5m2 + 2m

When r = 2

a = 5m + 2

Squaring both hand side , we get

a2 = ( 5m + 2 )2

a2 = 25m2 + 5 + 20m

a2 = 5 ( 5m2 + 4m + 5)

a2 = 5q , Where q = 5m2 + 5m + 1

When r = 3

a = 5m + 3

Squaring both hand side , we get

a2 = ( 5m + 3 )2

a2 = 25m2 + 9 + 30m

a2 = 25m2 + 30m + 10- 1

a2 = 5 ( 5m2 + 6m + 2 ) - 1

a2 = 5q -1 , where q = 5m2 + 6m + 2

Hence

Square of any positive integer is in form of 5q or 5q + 4. , where q is any integer . ( Hence proved )

3.Let a be any odd positive integer with b=4.

Therefore,

a= 4q+1 ( By Euclid's Division Lemma)

When a=4q+1

a2=(4q+1)2

=16q2 +1 + 8q

=8(2q2 +q) +1

=8q+1 where (q=2q2+q)

Hence proved.

4.Let a be a given integer.

On dividing a by 6 , we get q as the quotient and r as the remainder such that

a = 6q + r, r = 0,1,2,3,4,5

when r=0

a = 6q,even no

when r=1

a = 6q + 1, odd no

when r=2

a = 6q + 2, even no

when r = 3

a=6q + 3,odd no

when r=4

a=6q + 4,even no

when r=5,

a= 6q + 5 , odd no

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

Hope this helps you...

Please mark it as brainliest answer...☺☺☺

Answer 2

Answer:

D.All of the above

Step-by-step explanation:

Because we know that any positive integer is of the form a=bq+r where r is equal to or greater than 0 but smaller than b.

So we have a positive integer a

b=4 ,q=m and r = 0,1,2,3( why? mentioned above)

so every positive integer is of all of the forms given.

i.e 4m,4m+1,4m+2 and 4m+3.

Hope it Helps..