Question

show that any odd positive integer is of the form 4q+1 or 4 q +3 where q is some whole number

Answer 1

Hey there!

Let a be any positive integer

We know by Euclid's algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying, a = bq + r where 0 ≤ r < b.

Take b = 4

a = 4q + r

Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.

That is, can be , 4q or 4q + 1 or 4q + 2 or 4q + 3 where q is the quotient.

Since a is odd, a cannot be 4q or 4q + 2 as they are both divisible by 2.

Therefore, any odd integer is of the form 4q + 1 or 4q + 3.



Hope it help


Hope it help

Answer 2

$\huge \bf{ \red{hey}}$

$\huge{ \mathfrak{ \blue{here \: is \: answer}}}$

let a be any positive integer

then

b= 4

a= bq+r

0≤r<b

0≤r<4

r= 0,1,2,3

case 1.

r=0

a= bq+r

4q+0

4q

case 2.

r=1

a= 4q+1

6q+1

case3.

r=2

a=4q+2

case 4.

r=3

a=4q+3

hence from above it is proved that any positive integer is of the form 4q, 4q+1,4q+2,4q+3

$\huge \boxed{ \boxed{ \green{HOPE\: IT \: HELPS}}}$

$\huge{ \pink{thanks}}$