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

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

**(i) 1/âˆš2**

**Solution:**

Let assume that 1/âˆš2 is a rational numberLet us assume 1/âˆš2 = r where r is a rational number1/r = âˆš2We have assume that r is a rational number, 1/r = âˆš2 is also a rational numberBut as we know that âˆš2 is an irrational numberSo what we have assume is wrong.So we can say that, 1/âˆš2 is an irrational number.

**(ii) 7âˆš5**

**Solution:**

Letâ€™s assume that 7âˆš5 is a rational number.Again assume that two positive integers a and b.7âˆš5 = a/b here a and b are co-primesâ‡’ âˆš5 = a/7bâ‡’ âˆš5 is rational [ a and b are integers â‡’ a/7b is a rational number]This shows that âˆš5 is irrational. So, our assumption is wrong.So we can say that 7âˆš5 is an irrational number.

**(iii) 6 + âˆš2**

**Solution:**

Letâ€™s assume on that 6+âˆš2 is a rational number.Then, there are co prime positive integers a and b6 + âˆš2 = a/bâ‡’ âˆš2 = a/b â€“ 6â‡’ âˆš2 = (a â€“ 6b)/bâ‡’ âˆš2 is rational [(a-6b)/b is a rational number]This contradicts that âˆš2 is irrational. So, our assumption is incorrect.So we can say that 6 + âˆš2 is an irrational number.

**(iv) 3 âˆ’ âˆš5**

**Solution:**

Letâ€™s assume on that 3-âˆš5 is a rational number.There exist co prime positive integers a and b such that3-âˆš5 = a/bâ‡’ âˆš5 = a/b + 3â‡’ âˆš5 = (a + 3b)/bâ‡’ âˆš5 is rational [(a+3b)/b is a rational number]This contradicts that âˆš5 is irrational. our assumption is incorrect.So we can say that 3-âˆš5 is an irrational number.

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

**(i) 2/âˆš7**

**Solution:**

Letâ€™s assume that 2/âˆš7 is a rational number.There exist co-prime positive integers a and b2/âˆš7 = a/bâ‡’ âˆš7 = 2b/aâ‡’ âˆš7 is rational [2b/a is a rational number]This contradicts that âˆš7 is irrational. So, we can say that our assumption is incorrect.So we can say that, 2/âˆš7 is an irrational number.

**(ii) 3/(2âˆš5)**

**Solution:**

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etâ€™s assume that 3/(2âˆš5) is a rational number.There exist co â€“ prime positive integers a and b3/(2âˆš5) = a/bâ‡’ âˆš5 = 3b/2aâ‡’ âˆš5 is rational [3b/2a is a rational number]This contradicts that âˆš5 is irrational. So, our assumption is incorrect.So we can say that, 3/(2âˆš5) is an irrational number.

**(iii) 4 + âˆš2**

**Solution:**

Letâ€™s assume on the contrary that 4 + âˆš2 is a rational number.There exist co prime positive integers a and b4 + âˆš2 = a/bâ‡’ âˆš2 = a/b â€“ 4â‡’ âˆš2 = (a â€“ 4b)/bâ‡’ âˆš2 is rational [(a â€“ 4b)/b is a rational number]This contradict that âˆš2 is irrational. So, our assumption is incorrect.So we can say that 4 + âˆš2 is an irrational number.

**(iv) 5âˆš2**

**Solution:**

Letâ€™s assume on that 5âˆš2 is a rational number.There exist positive integers a and b such that5âˆš2 = a/b where, a and b, are co-primesâ‡’ âˆš2 = a/5bâ‡’ âˆš2 is rational [a/5b is a rational number]This contradicts that âˆš2 is irrational. So, our assumption is incorrect.So we can say that, 5âˆš2 is an irrational number.

**Question 3. Show that 2 âˆ’ âˆš3 is an irrational number.**

**Solution:**

Letâ€™s assume that 2 â€“ âˆš3 is a rational number.There exist co prime positive integers a and b2 â€“ âˆš3= a/bâ‡’ âˆš3 = 2 â€“ a/bâ‡’ âˆš3 = (2b â€“ a)/bâ‡’ âˆš3 is rational [(2b â€“ a)/b is a rational number]This contradicts that âˆš3 is irrational. So, our assumption is incorrect.So we can say that, 2 â€“ âˆš3 is an irrational number.

**Question 4. Show that 3 + âˆš2 is an irrational number.**

**Solution:**

Letâ€™s assume on that 3 + âˆš2 is a rational number.There exist co prime positive integers a and b3 + âˆš2= a/bâ‡’ âˆš2 = a/b â€“ 3â‡’ âˆš2 = (a â€“ 3b)/bâ‡’ âˆš2 is rational [(a â€“ 3b)/b is a rational number]This contradicts that âˆš2 is irrational. So, our assumption is incorrect.So we can say that, 3 + âˆš2 is an irrational number.

**Question 5. Prove that 4 âˆ’ 5âˆš2 is an irrational number.**

**Solution:**

Letâ€™s assume that 4 â€“ 5âˆš2 is a rational number.There exist co prime positive integers a and b4 â€“ 5âˆš2 = a/bâ‡’ 5âˆš2 = 4 â€“ a/bâ‡’ âˆš2 = (4b â€“ a)/(5b)â‡’ âˆš2 is rational [(4b â€“ a)/5b is a rational number]This contradicts that âˆš2 is irrational. So, our assumption is wrong.So we can say that, 4 â€“ 5âˆš2 is an irrational number.

**Question 6. Show that 5 âˆ’ 2âˆš3 is an irrational number.**

**Solution:**

Letâ€™s assume on that 5 â€“ 2âˆš3 is a rational number.There exist co prime positive integers a and b5 â€“ 2âˆš3 = a/bâ‡’ 2âˆš3 = 5 â€“ a/bâ‡’ âˆš3 = (5b â€“ a)/(2b)â‡’ âˆš3 is rational [(5b â€“ a)/2b is a rational number]This contradicts that âˆš3 is irrational. So, our assumption is wrong.So we can say that, 5 â€“ 2âˆš3 is an irrational number.

**Question 7. Prove that 2âˆš3 âˆ’ 1 is an irrational number.**

**Solution:**

Letâ€™s assume that 2âˆš3 â€“ 1 is a rational number.There exist co prime positive integers a and b2âˆš3 â€“ 1 = a/bâ‡’ 2âˆš3 = a/b + 1â‡’ âˆš3 = (a + b)/(2b)â‡’ âˆš3 is rational [(a + b)/2b is a rational number]This contradicts that âˆš3 is irrational. So, our assumption is wrong.So we can say that, 2âˆš3 â€“ 1 is an irrational number.

**Question 8. Prove that 2 âˆ’ 3âˆš5 is an irrational number.**

**Solution:**

Letâ€™s assume on that 2 â€“ 3âˆš5 is a rational number.There exist co prime positive integers a and b such that2 â€“ 3âˆš5 = a/bâ‡’ 3âˆš5 = 2 â€“ a/bâ‡’ âˆš5 = (2b â€“ a)/(3b)â‡’ âˆš5 is rational [(2b â€“ a)/3b is a rational number]This contradicts that âˆš5 is irrational. So, our assumption is wrong.So we can say that, 2 â€“ 3âˆš5 is an irrational number.

**Question 9. Prove that âˆš5 + âˆš3 is irrational.**

**Solution:**

Letâ€™s assume on that âˆš5 + âˆš3 is a rational number.There exist co prime positive integers a and bâˆš5 + âˆš3 = a/bâ‡’ âˆš5 = (a/b) â€“ âˆš3â‡’ (âˆš5)^{2}= ((a/b) â€“ âˆš3)^{2}[Squaring on both sides]â‡’ 5 = (a^{2}/b^{2}) + 3 â€“ (2âˆš3a/b)â‡’ (a^{2}/b^{2}) â€“ 2 = (2âˆš3a/b)â‡’ (a/b) â€“ (2b/a) = 2âˆš3â‡’ (a^{2}â€“ 2b^{2})/2ab = âˆš3â‡’ âˆš3 is rational [(a^{2}â€“ 2b^{2})/2ab is rational]This contradicts that âˆš3 is irrational. So, our assumption is wrong.so we can say that, âˆš5 + âˆš3 is an irrational number.

**Question 10. Prove that âˆš2 + âˆš3 is irrational.**

**Solution:**

Letâ€™s assume on that âˆš2 + âˆš3 is a rational number.There exist co prime positive integers a and b.âˆš2 + âˆš3 = a/bâ‡’ âˆš2 = (a/b) â€“ âˆš3â‡’ (âˆš2)^{2}= ((a/b) â€“ âˆš3)^{2}[Squaring on both sides]â‡’ 2 = (a^{2}/b^{2}) + 3 â€“ (2âˆš3a/b)â‡’ (a^{2}/b^{2}) + 1 = (2âˆš3a/b)â‡’ (a/b) + (b/a) = 2âˆš3â‡’ (a^{2}+ b^{2})/2ab = âˆš3â‡’ âˆš3 is rational [(a^{2}+ 2b^{2})/2ab is rational]This contradicts that âˆš3 is irrational. So, our assumption is wrong.So we can say that, âˆš2 + âˆš3 is an irrational number.

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

**Solution:**

Assume that âˆšp as a rational numberAgain Assume that âˆšp = a/b where a and b are integers and b â‰ 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, âˆšp is an irrational number.

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

**Solution:**

Letâ€™s assume on the contrary that âˆšp + âˆšq is a rational number.Then, there exist co prime positive integers a and b such thatâˆšp + âˆšq = a/bâ‡’ âˆšp = (a/b) â€“ âˆšqâ‡’ (âˆšp)^{2}= ((a/b) â€“ âˆšq)^{2}[Squaring on both sides]â‡’ p = (a^{2}/b^{2}) + q â€“ (2âˆšq a/b)â‡’ (a^{2}/b^{2}) â€“ (p+q) = (2âˆšq a/b)â‡’ (a/b) â€“ ((p+q)b/a) = 2âˆšqâ‡’ (a^{2}â€“ b^{2}(p+q))/2ab = âˆšqâ‡’ âˆšq is rational [(a^{2}â€“ b^{2}(p+q))/2ab is rational]This contradicts that âˆšq is irrational. So, our assumption is wrong.so we can say that, âˆšp + âˆšq is an irrational number.