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Question 1. Asked on :19 April 2019:08:34:46 AM

#### Given that HCF (306,657) =9 find LCM (306,657).

-Added by Prince Rawat Mathematics » Real Number

Himanshi Verma
Product of LCM and HCF= Product of two numbers

9×LCM = 306×657

9×LCM = 201042

LCM   = 22338

-Answered by Himanshi Verma On 21 April 2019:10:32:51 AM

Question 2. Asked on :19 April 2019:08:21:00 AM

#### Prove that√5 is irrational.

-Added by Prince Rawat Mathematics » Real Number

Himanshi Verma

Let us assume that √5 is rational

Then √5 =

(a and b are co primes, with only 1 common factor and b≠0)

⇒ √5 =

(cross multiply)

⇒ a = √5b

⇒ a² = 5b² -------> α

⇒ 5/a²

(by theorem if p divides q then p can also divide q²)

⇒ 5/a ----> 1

⇒ a = 5c

(squaring on both sides)

⇒ a² = 25c² ----> β

From equations α and β

⇒ 5b² = 25c²

⇒ b² = 5c²

⇒ 5/b²

(again by theorem)

⇒ 5/b---> 2

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.

This contradiction arises because we assumed that √5 is a rational number

∴ our assumption is wrong

∴ √5 is irrational number.

-Answered by Himanshi Verma On 21 April 2019:11:24:50 AM

Question 3. Asked on :19 April 2019:08:18:18 AM

#### Check whether 6n can end digit 0 for any nature number n.

-Added by Prince Rawat Mathematics » Real Number

Akki chauhan

prime factors of 6n

= (2*3)n

= 2n*3n

While the number ending with zero have prime factor as 2n*5n

Hence, 6n is not end with zero.

-Answered by Akki chauhan On 28 June 2019:09:30:20 AM

Question 4. Asked on :18 April 2019:08:18:35 AM

#### Use euclid's division alogrithan to find the HCF of :(i) 135 and 225 .

-Added by Nitish kumar Mathematics » Real Number

Himanshi Verma

Use Euclid division alogrithan

a=bq+r

225=135×1+90

135=90×1+45

90=45×2+0

HCF=45 .

-Answered by Himanshi Verma On 18 April 2019:04:26:31 PM

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