# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 are part ofÂ NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

• Real Numbers Class 10 Ex 1.1
• Real Numbers Class 10 Ex 1.2
• Real Numbers Class 10 Ex 1.3
• Real Numbers Class 10 Ex 1.4

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Ex 1.3 Class 10 Maths Question 1.
Prove that âˆš5 is irrational.
Solution:
Let âˆš5 =Â $\frac { p }{ q }$Â be a rational number, where p and q are co-primes and q â‰  0.
Then, âˆš5q = p => 5q2=p2
â‡’Â  p2Â â€“ Sq2Â  Â  Â â€¦ (i)
Since 5 divides p2, so it will divide p also.
Let p = 5r
Then p2Â â€“ 25rÂ 2Â  Â  Â [Squaring both sides]
â‡’ 5q2Â = 25r2Â  Â  Â [From(i)]
â‡’ q2Â = 5r2
Since 5 divides q2, so it will divide q also. Thus, 5 is a common factor of both p and q.
This contradicts our assumption that âˆš5 is rational.
Hence, âˆš5 is irrational. Hence, proved.

Ex 1.3 Class 10 Maths Question 2.
Show that 3 + âˆš5 is irrational.
Solution:
Let 3 + 2âˆš5 = $\frac { p }{ q }$ be a rational number, where p and q are co-prime and q â‰  0.
Then, 2âˆš5 = $\frac { p }{ q }$ â€“ 3 = $\frac { p - 3q }{ q }$
â‡’ âˆš5 =  $\frac { p - 3q }{ 2q }$
since  $\frac { p - 3q }{ 2q }$ is a rational number,
therefore, âˆš5 is a rational number. But, it is a contradiction.
Hence, 3 + âˆš5 is irrational. Hence, proved.

Ex 1.3 Class 10 Maths Question 3.
Prove that the following are irrational.

Solution:
(i) Let  $\frac { 1 }{ \sqrt { 2 } }$ = $\frac { p }{ q }$ be a rational number,
where p and q are co-prime and q â‰  0.
Then, âˆš2 = $\frac { q }{ p }$
Since $\frac { q }{ p }$ is rational, therefore, âˆš2 is rational.
But, it is a contradiction that âˆš2 is rational, rather it is irrational.
Hence, $\frac { 1 }{ \sqrt { 2 } }$ is irrational.
Hence, proved.

(ii) Let 7âˆš5 = $\frac { p }{ q }$ be a rational number, where p, q are co-primes and q â‰  0.
Then, âˆš5 = $\frac { p }{ 7q }$
Since $\frac { p }{ 7q }$ is rational therefore, âˆš5 is rational.
But, it is a contradiction that âˆš5 is rational rather it is irrational.
Hence, 7âˆš5 s is irrational.
Hence proved.

(iii) Let 6 + âˆš2 = $\frac { p }{ q }$ be a rational number, where p, q are co-primes and q â‰  0.
Then, âˆš2 = $\frac { p }{ q }$ â€“ 6 = $\frac { p - 6q }{ q }$
Since $\frac { p - 6q }{ q }$ is rational therefore, âˆš2 is rational.
But, it is a contradiction that âˆš2 is rational, rather it is irrational.
Hence, 6 + âˆš2 is irrational.

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