1.3 Revisiting Irrational Numbers:
A number‘s’ is called irrational if it cannot be written in the form p /q
where p and q are integers and q ≠ 0.
Theorem 1.3: Let p be a prime number. If p divides a2 , then p divides a, where a is a positive integer
Proof: Let the prime factorisation of a be as follows:
a = p1 p2 . . . pn , where p1 p2, ……. ., pn are primes,
Therefore, a2 = (p1 p2 . . . pn ) ( p1p2 . . .pn ) = { p }_{ 1 }^{ 2 },{ p }_{ 2 }^{ 2 }..........{ p }_{ n }^{ 2 }
Since p divides a2
p is one of the prime factors of a2 .( By Fundamental Theorem of Arithmetic ).
Also the only prime factors of a2 are p1 p2 . . . pn. So p is one of p1 p2 . . . pn
Since a = p1 p2 . . . pn , p divides a.
Theorem 1.4: Prove that √2 is irrational
Proof: Let us assume, that √2 is rational. So, we can find integers r and s (≠ 0) such that √2 = r/s Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get √2 = a/b , where a and b are coprime. So, b√2 = a ∴ 2b2 = a2 (squaring on both sides)
2 = a2/b2
Now, by Theorem 1.3, it follows that 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2. Hence 2 divides b2, and so 2 divides b
Therefore, a and b have at least 2 as a common factor, contradicting the fact that a and b have no common factor other than 1. So, we conclude that is irrational.
Example: Prove that √3 is irrational.
Proof: Let us assume, to the contrary, that √3 is rational.
We can find integers a and b (≠0) such that √3 = \frac { a }{ b }
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are co prime. So b √3 = a We get 3 b² = a². (Squaring on both sides) Hence, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.
So , a = 3c for some integer c. Substituting for a, we get 3b²= 9c², b² = 3c². Now b² is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3) Therefore, a and b have at least 3 as a common factor. Contradicting the fact that a and b are coprime. So, we conclude that √3 is irrational.
The sum or difference of a rational and an irrational number is irrational.
The product and quotient of a non-zero rational and irrational number is irrational.
Example: Show that 5 √3 is irrational
Proof: Let us assume, to the contrary, that 5 √3 is rational we can find coprime a and b (b ≠ 0) such that 5 √3 =a/b Therefore, 5 – ( a)/b = √3
Rearranging this equation, we get √3 = 5 a/b = (5b − a)/b Since a and b are integers, we get 5 a/b is rational, and so √3 is rational contradicting to the fact that √3 is irrational. So, we conclude that 5 √3 is rational
Example: Show that 3 √2 is irrational.
Proof: Let us assume, to the contrary, that 3 √2 is rational. we can find coprime a and b (b ≠ 0) such that 3 √2 = a/b
Rearranging, we get √2 = a/3b
Since 3, a and b are integers, a/3b is rational and so √2 is rational But this contradicts the fact that√2 is irrational. So, we conclude that 3 √2 is irrational.
EXERCISE 1.3
1: Prove that √5 is irrational.
Sol: We will prove this using contradiction method
Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = \frac { a }{ b }
Multiply by b both sides we get
√5 x b = a
To remove root, squaring on both sides, we get
5b2 = a2 … ( I )
Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which divides a2 then it will divide a also.
That means 5 will divide a. So we can write a = 5k
Putting value of a in equation ( I ) we get
5b2 = (5k)2 , 5b2 = 25 k2
Dividing by 5 we get b2 = 5k2
Similarly, we get that b will be divisible by 5
and we have already got that a is divisible by 5
but a and b are co-prime number. So it contradicts.
Hence √5 is not a rational number, it is irrational.
2. Prove that the following are irrationals:
(i) 1/(√2) (ii) 7√5 (iii) 6 + √2
Sol: (i) Let take that 1/(√2) is a rational number.
So we can write this number as (1 )/(√2) = \frac { a }{ b }
Here a and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get 1 = (a√2)/b)
Now multiply by b we get b = a√2
dividing by a we get ( b)/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number
But √2 is a irrational number so it contradicts.
Hence, 1/(√2) is a irrational number
(ii) Let take that 7√5 is a rational number.
So we can write this number as 7√5 = \frac { a }{ b }
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get √5 = (a/7b)
Here a and b are integer so (a/7b) is a rational number So √5 should be a rational
Number but √5 is a irrational number so it contradicts.
Hence, 7√5 is a irrational number.
(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = (\frac { a }{ b } )
Here a and b are two co prime number and b is not equal to 0
Subtracting 6 from both sides we get
√2 = (\frac { a }{ b } ) – 6
√2 = ((a−6b)/b)
Here a and b are integer so (a−6b)/b) is a rational number so √2 should be a rational number.
But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.
Additional Practice Questions:
Q1. Prove that √7 is irrational.
Q2. Prove that 7 + 3 √5 is irrational.
Q3.Prove that the following are irrationals:
(a) 1/√5 (b) 3√5 (c) 9 + √3