Proof
We will prove this by contradiction.
Let us add a rational number ( ), to an irrational number ( = p ). We assume that the result is a rational number ( ).
Putting these in an equation, we get:
We can express p as a rational number. This result contradicts the fact that it is an irrational number. Therefore our assumption (sum of rational and irrational numbers is a rational number) is incorrect. So this sum is an irrational number. This result holds for difference also.
✩ Known Irrationals
- Square Root of Prime (): √2, √3, √5, √7, √11, √13, √17, √19 …
- Special Numbers: Pi ( π ) , Euler’s number ( e ), Golden Ratio
- Logarithms of primes with prime base: log23, log35…
Examples of Addition
- 4 + √3 is Irrational, as 4 is rational and √3 is irrational.
- 6 – √2 is Irrational, as 6 is rational and √2 is irrational.
- π – 2 is Irrational, as π is irrational and 2 is rational.
- √7 – 8 is Irrational, as √7 is irrational and 8 is rational.
✩ Irrational Result of Operations
Following operations between rational and irrational numbers result in an irrational number. Whatever the order of operations, the outcome is always an irrational number.
- Rational + Irrational: [ 3 + √2 ], [ 4 + √7 ], …
- Rational − Irrational: [ 5 – √2 ], [ √3 – 6 ], …
- Rational × Irrational: [ 4 × π = 4π ], [ 6 × √3 = 6√3 ], …
- Rational ÷ Irrational: [ 2 ÷ √2 ], [ π ÷ 2 ], …
Sum of two Irrational numbers
What happens when we add two irrational numbers? In this case, the resulting number may be rational or irrational. Let us see some examples.
Examples
In the following examples a and b are irrational numbers.
1. Example
a = (2 + √5), b = (3 − √5)
We find the sum a + b:
a + b = (2 + √5) + (3 − √5) = 5
Result is a rational number.
2. Example
a = π, b = 6 + π
Let us find the sum of a and b:
a + b = π + (6 + π) = 6 + 2π
Result is an irrational number.
Related
Examples of Irrational Numbers (With Lists) ➤
Square root of a prime number is Irrational ➤
Product of rational & irrational numbers is irrational: Proof, Examples ➤