Let  is rational

i.e., it can be expressed as whereas ‘a’ and ‘b’ both are integers and b ≠ 0.
Thus,  



Now  is rational and w'e know that 2 is also rational.
 

is also rational

[∵ Difference, sum and product aftwo rational numbers are always rational]
Comparing it with result (i), we get  is rational, which is not true as  is an irrational number.

∴ Our assumption that     is rational is not correct.

In order to arrange the books as required, we have to find the largest number that divides 12, 20 and30 exactly. Clearly, we have to find HCF. Now,

Since, 12 = 2 × 2 × 3 = 2² × 3

20 = 2 × 2 × 5 = 2² × 5

and    30 = 2 × 3 × 5

HCF = Product of common prime factors with lowest powers

= 2

So, there must be 2 books in each stack.

∴ Number of stacks of Physics books


Number of stacks of Chemistry books



Number of stacks of Mathematics books

Let us assume that   is is rational. We know that, rational number can be written as   where ‘a’ and ‘b’ are integers and b ≠ 0. i.e.. Assume that



Squaring both sides

⇒ a² is divisible by 2

⇒ a is divisible by 2

Let a = 2c for some integer c.

Putting a = 2c in (i), we get

2 b² = (2c)²

⇒    2b² = 4c²

⇒    b² — 2c²

⇒ b² is divisible by 2

⇒ b is divisible by 2.

Thus, 2 is a common factor of a and b. But, this contradicts the fact that ‘a’ and ‘b’ have no common factor other than 1.

The contradiction arises by assuming that   is rational.
Hence,   is irrational.

Required distance is the LCM of 84 cm, 90 cm and 120 cm.

Thus,

Since, 84 = 2 × 2 × 3 × 7 = 2² × 3 × 7

90 = 2 × 3 × 3 × 5 = 2 × 3² × 5

and    120 = 2 × 2 × 2 × 3 × 5

= 2³ × 3 × 5

∴ LCM = Product of each prime factor with highest power

= 2³ × 3² × 5 × 7 = 2520

= 25 m 20 cm

Problems Based on Irrational Numbers

Let us assume that  is rational.

We know that rational number can be written as  where ‘a’ and ‘b’ are integers and b ≠ 0.
i.e., assume that



Squaring both side, we get

⇒ a² is divisible by 3

⇒ a is divisible by 3

Let a = 3c for some integer ‘c’.

Putting a = 3c in (i)

a² = 3b²

⇒    (3c)² = 3b²

⇒    9c² = 3b²

⇒    b² = 3c²

⇒ b² is divisible by

⇒ b is divisible by 3

Thus, 3 is a common factor of ‘a’ and ‘b’.

But this contradicts the fact that ‘a’ and ‘b’ have no common factor other than 1.
The contradiction arises by assuming that  is rational.
Hence,    is irrational.