Question

use euclids divisio lemma to show that any positive odd intiger is of the form 6q +1, or 6q+ 3 or 6q + 5 , where q is some integers ​

Answer 1

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✤ Required Proof:-

✒ To prove:

• Show that any positive integer is of the form 6q + 1, 6q + 3 or 6q + 5, where q is some integer.

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✤ How to solve?

First of all, Let's know what is Euclid's division lemma:

For any positive integers a and b, there exist unique integers q and r such that,

$\large{ \boxed{ \sf{a = bq + r}}}$

Where, 0 $\sf{\leqslant}$r < b

Or simply,

Dividend = Divisor × Quoteint + Remainder

• Quotient is some positive numbers when a is greater or equal to b.
• Quotient = 0 when a < b
• Remainder is a positive no. or 0, Whose value is less than Divisor (b).

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✤ Solution:-

Let a be any +ve integer and b = 8(Consider)

By using Euclid's division lemma,

$\large{ \sf{ \longrightarrow \: a = 6q + r}}$

Where, q is any quotient we get and r is the remainder.

So, possible reminders are:

➝ More than or equal to 0 but less than 6 [ 0, 1, 2, 3, 4, 5]

So, According to our equation, a =

➝ 6q = 2(3q)

➝ 6q + 1 = 2(3q) + 1

➝ 6q + 2 = 2(3q + 1)

➝ 6q + 3 = 2(3q + 1) + 1

➝ 6q + 4 = 2(3q + 2)

➝ 6q + 5 = 2(3q + 2) + 1

So, Here 6q, 6q + 2 and 6q + 4 are divisible by 2, hence they are not odd integers. So, odd positive integers are 6q + 1, 6q + 3, 6q + 5.

☃️ Hence, solved !!

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Answer 2

☬ $\purple{\mathbb{ANSWER}}$ ☬

➾ EUCLID'S DIVISION LEMMA:—

➲ For any positive integers a and b, there exist unique integers q and r such that:–

a= bq + r and 0 ≤ r < b

➾ SOLUTION:—

➲ Let b = 6 and so values of r = 0,1,2,3,4&5.

∴ Any positive integer a :–

☼a = 6q + 0 ∴ a = 6q

☼a = 6q + 1

☼a = 6q + 2

☼a = 6q + 3

☼a = 6q + 4

☼a = 6q + 5

➾ YOUR ANSWER:—

❶ a = 6q + 1

❷ a = 6q + 3

❸ a = 6q + 5.

Hence, Proved.