Question

show that any positive even integer is of the form 4q or 4q+2 and any positive odd integer is of the form 4q+1 or 4q+3 where q is any integer.​

Answer 1

1) Show that any positive even integer is of the form 4q or 4q+2.

Let a and b be any positive even integer. (a > b)

Then by Euclid's division lemma.

a = bq + r

Take b = 4 (0 ≤ r < b)

a = 4q + r

Where r = 0, 1, 2, 3

When r = 0

a = 4q + (0)

a = 4q = 2(2q) is an even number

When r = 2

a = 4q + (2)

Take 2 as common

a = 2(2q + 1) which is also an even number

So, we can say that any positive even integer is in the form 4q or 4q + 2 where q is any integer.

2) Show that any positive odd integer is of the form 4q+1 or 4q+3 where q is any integer.

Let a and b be any odd integer. (a > b)

Then by Euclid's division lemma.

a = bq + r

Take b = 4 (0 ≤ r < b)

a = 4q + r

Where r = 0, 1, 2, 3

When r = 1

a = 4q + 1 is a odd number

When r = 3

a = 4q + 3 which is also an odd number

So, we can say that any odd integer is in the form 4q + 1 or 4q + 3 where q is any integer.

Answer 2

Question (1) :- show that any positive even integer is of the form 4q or 4q+2 ?

Answer :-

Euclid's Division Lemma :- states that, if two positive integers “a” and “b”, then there exists unique integers “q” and “r” such that which satisfies the condition

☛ a = bq + r where 0 ≤ r ≤ b.

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Let a be the positive integer, And b = 4 .

Then by Euclid's division lemma, We can write a = 4q + r , for some integer q and 0 ≤ r < 4 .

Then, possible values of r is 0, 1, 2 and 3 .

Taking r = 0

→ a = 4q

Taking r = 1

→ a = 4q + 1

Taking r = 2

→ a = 4q + 2

Taking r = 3

→ a = 4q + 3

But a is an even positive integer, so a can't be 4q + 1 , or 4q + 3 [ As these are odd ] .

Hence, any Even integer can be of the form 4q or 4q + 2 for some integer q .

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Question (2) :- show that any positive odd integer is of the form 4q+1 or 4q+3 where q is any integer. ?

Answer :-

Let a any odd positive integer and b = 4..

Using Euclid Division Lemma, a = 4q + r, where and 0 r ≤ 4..

So, possible values of r is 0, 1, 2 and 3 .

Taking r = 0

→ a = 4q

Taking r = 1

→ a = 4q + 1

Taking r = 2

→ a = 4q + 2

Taking r = 3

→ a = 4q + 3

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are Even ] .

Hence, any odd integer can be of the form (4q+1) or (4q + 3) for some integer q .

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