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
Prove that is irrational.
⇒ 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
Show that is an irrational number.
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.
Prove that is irrational.
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.