Proof
We prove this by contradiction.
Let be a rational number and p an irrational number.
Let us assume that product of these numbers is a rational number . We are hoping to get a contradiction due to this assumption. The assumption results in the following equation:
Multiplying both sides by :
The equation expresses as a product of two rational numbers. So it must be rational. This result contradicts the fact that is an irrational number.
So our assumption ( product of a rational number with an irrational number p is a rational number ) is false. Therefore the result of this product is an irrational number.
✩ 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
The outcome of multiplication in the following examples is an irrational number. Why?
One of the numbers in multiplication is a square root of prime number, which is irrational.
Example 1
- 2 is rational
- √3 is irrational
- Product is irrational
Example 2
- is rational
- √5 is irrational
- Product is irrational
Example 3
is rational, is irrational
Their product is irrational.
Example 4
Following products are irrational because π is an irrational number.
- 2 × π = 2π
✩ 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 ], …
- Non zero rational × Irrational: [ 4 × π = 4π ], [ 6 × √3 = 6√3 ], …
- Non zero rational ÷ Irrational: [ 2 ÷ √2 ], [ π ÷ 2 ], …
Product of two irrational numbers
We know that product of two rational numbers is rational. The only remaining case is the product of two irrational numbers. In this case, the resulting number may be rational or irrational, depending on the multiplicand and multiplier.
Let us see some examples.
Example 5
√2 × √6
= 2√3
This product is an irrational number due to the presence of the square root of √3, which is an irrational number.
Example 6
√3 × √3
= √9
= 3
3 is a rational number. The product of two irrational numbers can be rational.
Example 7
(2 + √3) × (2 − √3)
The resulting number in each bracket above is a sum of a rational and an irrational number. So it is an irrational number. Let us see what happens to the product.
(2 + √3) × (2 − √3)
= 4 − 2√3 + 2√3 − (√3)2
= 4 − 3 = 1
The product is a rational number!
Questions
1 Question
is rational or irrational?
Answer
Is 31 a prime number?
2 Question
is rational or irrational?
Answer
Can you multiply the numbers?
3 Question
Find the product (4 − √7)(4 + √7). Is it rational?
Answer
Can you apply the formula (a − b)(a + b) = a2 − b2?
4 Question
Prove that the area of circle with radius is a rational number.
Answer
Can you find the area using the formula Area = πr2?
5 Question
Given , where x, y are rational numbers. Find x, y.
Answer
Can you rationalize the denominator?
Answers
1 Answer
31 is a prime number.
Square root of prime is irrational, so its product with 2 is also irrational.
2 Answer
On multiplication we get:
4√5 × √5 = 4 × 5 = 20
This is a rational number.
3 Answer
Applying the formula for difference of squares, we get:
(4 − √7)(4 + √7) = 42 − (√7)2
= 16 − 7 = 9
The result is a rational number.
4 Answer
The area of a circle = πr2. Applying this formula:
Area
The result is a rational number.
5 Answer
Let us rationalize the denominator by multiplying the left side of the equation with :
−9 − 2√5 = x + y√5
⇒ x = − 9, y = − 2
Related
Examples of Irrational Numbers (With Lists) ➤