Question

show that every integer is of the form 2q and positive odd integer 2q+1 where q is some integer​

Answer 1

AnswEr:

Let us consider that a & b are two odd positive Integers.

Then, By using Euclid's Division lemma.

$:\implies\tt\; a = bq + r , 0 < r$

Here, b = 2 and r is remainder and value of q is more than or Equal to zero, r can be 0 and 1 because 0 < r < b and the value of b is 2.

$:\implies\tt\;a = 2q$

$:\implies\tt\;a = 2q + 1$

$\therefore\tt\; a = 2q \; or \; a = 2q + 1$

So, Total possible forms are 2q and 2q + 1

$\rule{150}2$

If 2q + 0

Here 2 is divisible so it is an even number.

If a = 2q + 1

Here, 2 is divisible by 2 but 1 is not divisible by 2 so it is an odd number.

Integer can be even or odd.

Hence, any odd positive integer is in the form of 2q + 1.

$\rule{150}2$

Answer 2

Answer:

Euclid Division lemma:

⇒ As per Euclid division lemma, If a and b are two positive integers, then, a = bq + r.

Where, 0 ≤ r ≤ b.

$\rule{200}{2}$

$\sf Let\;positive\;integer\;be\;a\;and\;b=2,\\ \\ \sf Hence,\;a=2bq+r\\ \\ \sf Where,\;0\leq r<2,\;and\;r\;is\;an\;integer\;\leq 0\;and\;less\;than\;2.\\ \\ \sf Hence,\;r\;can\;be\;either\;1\;or\;0.$

$\rule{200}{2}$

$\bf Case-1).\;If\;r=0\\ \\ \implies \sf a=2q+r\\ \\ \implies \sf a=2q\\ \\ \underline{\sf This\;will\;always\;be\;an\;even\;integer.}\\ \\ \rule{200}{2}\\ \\ \bf Case-2).\;If\;r=1\\ \\ \implies \sf a=2q+r\\ \\ \implies \sf a=2q+1\\ \\ \underline{\sf This\;will\;always\;be\;an\;odd\;integer.}\\ \\ \rule{200}{2}$

$\underline{\bf Hence\;Proved!!}$