Question

show that any odd positive integer is of the form 4q + 1 or 4q+ 3​

Answer 1

QuesTion :-

Show that any odd positive integer is of the form 4q + 1 or 4q+ 3​

SoluTion :-

According to Euclid’s Division lemma,

If a and b are two positive integers

Then,

$\rm {a=bq+r}\\\\\\\rm {Where \ 0 \leq r < b.}$

Let the positive integers be a and b = 4

$\rm {\therefore \ a=bq+r}\\\\\\\rm {Where, (0 \leq r < 4)}$

So, r can be either 0, 1, 2 or 3

If r = 1 then,

$\rm {a=bq+r}\\\\\\\rm {a=4q+1}$

(Odd integer)

If r = 3 then,

$\rm {a=bq+r}\\\\\\\rm {a=4q+3}$

(Still an odd integer)

Hence proved.

Answer 2

Step-by-step explanation:

Question:-

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

We have

Any positive integer is of the form

$\red{ \bf \: 4q+1or \: 4q+3 \: }$

As per Euclid’s Division lemma.

As per Euclid’s Division lemma.If a and b are two positive integers, then,

$\bf \red{\: a=bq+r} \\ </strong></p><p></p><p><strong> \bf \red{\: a=bq+r}$

Where 0≤r<b.

Let positive integers be a.and b=4

Let positive integers be a.and b=4Hence,

$\bf \red{\: a=bq+r} \\ </p><p></p><p>$

Where,

(0≤r<4)

R is an integer greater than or equal to 0 and less than 4

Hence, r can be either 0,1,2and3

Hence, r can be either 0,1,2and3Now,

$\bf If \: r=1$

Then, our be equation is becomes

$\bf \: a=bq+r \\ </p><p></p><p> \bf \: a=4q+1$

This will always be odd integer.

Now, If r=3

Then, our be equation is becomes

$\bf \: a=bq+r \\ </p><p></p><p> \bf \: a=4q+3$

$\bf \pink{Hence \: Proved}$