# Class 10 RD Sharma Solutions- Chapter 1 Real Numbers – Exercise 1.5

**Question 1. Show that the following numbers are irrational.**

**(i) 1/√2**

**Solution:**

Let assume that 1/√2 is a rational numberLet us assume 1/√2 = r where r is a rational number1/r = √2We have assume that r is a rational number, 1/r = √2 is also a rational numberBut as we know that √2 is an irrational numberSo what we have assume is wrong.So we can say that, 1/√2 is an irrational number.

**(ii) 7√5**

**Solution:**

Let’s assume that 7√5 is a rational number.Again assume that two positive integers a and b.7√5 = a/b here a and b are co-primes⇒ √5 = a/7b⇒ √5 is rational [ a and b are integers ⇒ a/7b is a rational number]This shows that √5 is irrational. So, our assumption is wrong.So we can say that 7√5 is an irrational number.

**(iii) 6 + √2**

**Solution:**

Let’s assume on that 6+√2 is a rational number.Then, there are co prime positive integers a and b6 + √2 = a/b⇒ √2 = a/b – 6⇒ √2 = (a – 6b)/b⇒ √2 is rational [(a-6b)/b is a rational number]This contradicts that √2 is irrational. So, our assumption is incorrect.So we can say that 6 + √2 is an irrational number.

**(iv) 3 − √5**

**Solution:**

Let’s assume on that 3-√5 is a rational number.There exist co prime positive integers a and b such that3-√5 = a/b⇒ √5 = a/b + 3⇒ √5 = (a + 3b)/b⇒ √5 is rational [(a+3b)/b is a rational number]This contradicts that √5 is irrational. our assumption is incorrect.So we can say that 3-√5 is an irrational number.

**Question 2. Prove that the following numbers are irrationals.**

**(i) 2/√7**

**Solution:**

Let’s assume that 2/√7 is a rational number.There exist co-prime positive integers a and b2/√7 = a/b⇒ √7 = 2b/a⇒ √7 is rational [2b/a is a rational number]This contradicts that √7 is irrational. So, we can say that our assumption is incorrect.So we can say that, 2/√7 is an irrational number.

**(ii) 3/(2√5)**

**Solution:**

L

et’s assume that 3/(2√5) is a rational number.There exist co – prime positive integers a and b3/(2√5) = a/b⇒ √5 = 3b/2a⇒ √5 is rational [3b/2a is a rational number]This contradicts that √5 is irrational. So, our assumption is incorrect.So we can say that, 3/(2√5) is an irrational number.

**(iii) 4 + √2**

**Solution:**

Let’s assume on the contrary that 4 + √2 is a rational number.There exist co prime positive integers a and b4 + √2 = a/b⇒ √2 = a/b – 4⇒ √2 = (a – 4b)/b⇒ √2 is rational [(a – 4b)/b is a rational number]This contradict that √2 is irrational. So, our assumption is incorrect.So we can say that 4 + √2 is an irrational number.

**(iv) 5√2**

**Solution:**

Let’s assume on that 5√2 is a rational number.There exist positive integers a and b such that5√2 = a/b where, a and b, are co-primes⇒ √2 = a/5b⇒ √2 is rational [a/5b is a rational number]This contradicts that √2 is irrational. So, our assumption is incorrect.So we can say that, 5√2 is an irrational number.

**Question 3. Show that 2 − √3 is an irrational number.**

**Solution:**

Let’s assume that 2 – √3 is a rational number.There exist co prime positive integers a and b2 – √3= a/b⇒ √3 = 2 – a/b⇒ √3 = (2b – a)/b⇒ √3 is rational [(2b – a)/b is a rational number]This contradicts that √3 is irrational. So, our assumption is incorrect.So we can say that, 2 – √3 is an irrational number.

**Question 4. Show that 3 + √2 is an irrational number.**

**Solution:**

Let’s assume on that 3 + √2 is a rational number.There exist co prime positive integers a and b3 + √2= a/b⇒ √2 = a/b – 3⇒ √2 = (a – 3b)/b⇒ √2 is rational [(a – 3b)/b is a rational number]This contradicts that √2 is irrational. So, our assumption is incorrect.So we can say that, 3 + √2 is an irrational number.

**Question 5. Prove that 4 − 5√2 is an irrational number.**

**Solution:**

Let’s assume that 4 – 5√2 is a rational number.There exist co prime positive integers a and b4 – 5√2 = a/b⇒ 5√2 = 4 – a/b⇒ √2 = (4b – a)/(5b)⇒ √2 is rational [(4b – a)/5b is a rational number]This contradicts that √2 is irrational. So, our assumption is wrong.So we can say that, 4 – 5√2 is an irrational number.

**Question 6. Show that 5 − 2√3 is an irrational number.**

**Solution:**

Let’s assume on that 5 – 2√3 is a rational number.There exist co prime positive integers a and b5 – 2√3 = a/b⇒ 2√3 = 5 – a/b⇒ √3 = (5b – a)/(2b)⇒ √3 is rational [(5b – a)/2b is a rational number]This contradicts that √3 is irrational. So, our assumption is wrong.So we can say that, 5 – 2√3 is an irrational number.

**Question 7. Prove that 2√3 − 1 is an irrational number.**

**Solution:**

Let’s assume that 2√3 – 1 is a rational number.There exist co prime positive integers a and b2√3 – 1 = a/b⇒ 2√3 = a/b + 1⇒ √3 = (a + b)/(2b)⇒ √3 is rational [(a + b)/2b is a rational number]This contradicts that √3 is irrational. So, our assumption is wrong.So we can say that, 2√3 – 1 is an irrational number.

**Question 8. Prove that 2 − 3√5 is an irrational number.**

**Solution:**

Let’s assume on that 2 – 3√5 is a rational number.There exist co prime positive integers a and b such that2 – 3√5 = a/b⇒ 3√5 = 2 – a/b⇒ √5 = (2b – a)/(3b)⇒ √5 is rational [(2b – a)/3b is a rational number]This contradicts that √5 is irrational. So, our assumption is wrong.So we can say that, 2 – 3√5 is an irrational number.

**Question 9. Prove that √5 + √3 is irrational.**

**Solution:**

Let’s assume on that √5 + √3 is a rational number.There exist co prime positive integers a and b√5 + √3 = a/b⇒ √5 = (a/b) – √3⇒ (√5)^{2}= ((a/b) – √3)^{2}[Squaring on both sides]⇒ 5 = (a^{2}/b^{2}) + 3 – (2√3a/b)⇒ (a^{2}/b^{2}) – 2 = (2√3a/b)⇒ (a/b) – (2b/a) = 2√3⇒ (a^{2}– 2b^{2})/2ab = √3⇒ √3 is rational [(a^{2}– 2b^{2})/2ab is rational]This contradicts that √3 is irrational. So, our assumption is wrong.so we can say that, √5 + √3 is an irrational number.

**Question 10. Prove that √2 + √3 is irrational.**

**Solution:**

Let’s assume on that √2 + √3 is a rational number.There exist co prime positive integers a and b.√2 + √3 = a/b⇒ √2 = (a/b) – √3⇒ (√2)^{2}= ((a/b) – √3)^{2}[Squaring on both sides]⇒ 2 = (a^{2}/b^{2}) + 3 – (2√3a/b)⇒ (a^{2}/b^{2}) + 1 = (2√3a/b)⇒ (a/b) + (b/a) = 2√3⇒ (a^{2}+ b^{2})/2ab = √3⇒ √3 is rational [(a^{2}+ 2b^{2})/2ab is rational]This contradicts that √3 is irrational. So, our assumption is wrong.So we can say that, √2 + √3 is an irrational number.

**Question 11. Prove that for any prime positive integer p, √p is an irrational number.**

**Solution:**

Assume that √p as a rational numberAgain Assume that √p = a/b where a and b are integers and b ≠ 0By squaring on both sidesp = a^{2}/b^{2}pb = a^{2}/bp and b are integers pb= a^{2}/b will also be an integerBut we know that a^{2}/b is a rational number. so our assumption is wrongSo, √p is an irrational number.

**Question 12. If p, q are prime positive integers, prove that √p + √q is an irrational number.**

**Solution:**

Let’s assume on the contrary that √p + √q is a rational number.Then, there exist co prime positive integers a and b such that√p + √q = a/b⇒ √p = (a/b) – √q⇒ (√p)^{2}= ((a/b) – √q)^{2}[Squaring on both sides]⇒ p = (a^{2}/b^{2}) + q – (2√q a/b)⇒ (a^{2}/b^{2}) – (p+q) = (2√q a/b)⇒ (a/b) – ((p+q)b/a) = 2√q⇒ (a^{2}– b^{2}(p+q))/2ab = √q⇒ √q is rational [(a^{2}– b^{2}(p+q))/2ab is rational]This contradicts that √q is irrational. So, our assumption is wrong.so we can say that, √p + √q is an irrational number.