Question

show that every positive odd integer the form 4m or 4m +3 where m is a whole no.​

Answer 1

Answer:

Let a be any positive integer and b=4, then by Euclid's algorithm,

a=4q+r, for some integer q≥0 and r=0,1,2,3

So, a=4q or 4q+1 or 4q+2 or 4q+3 because 0≤r<4.

Now, 4q that is (2×2q) is an even number.

Therefore, 4q+1 is an odd number.

Now, 4q+2 that is 2(2q+1) which is also an even number.

Therefore, (4q+2)+1=4q+3 is an odd number.

Hence, we can say that any even integer can be written in the form of 4q or 4q+2 where q is a whole number.

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Step-by-step explanation:

Applying Euclids division algorithm with a,b,q and r where b=4 (Theorem 1.1)

⇒a=bq+r,0≤r<b

⇒a=4q+r,0≤r<4

i) When r=0,a=4q

Thus a

2

=16q

2

=4(4q

2

)=4Q where Q=4q

2

ii) When r=1,a=4q+1

⇒a

2

=(4q+1)

2

=16q

2

+6q+1=4q(4q+2)+1=4Q+1

where 4q+2=Q

iii) when r=2,a=4q+2

⇒a

2

=16q

2

+16q+4=4(4q

2

+4q+1)4R

Where a=4q

2

+4q+1

iv) When r=3,a=4q+3

⇒a

2

=(4q+3)

2

=16q

2

+24q+9=4(4q

2

+6q+2)+1

=42+1 when 2=4q

2

+q+1

Thus we can see that the square of any +ve integer is of the form 4Q or 4Q+1 for some integer Q

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