  # Prove that 6+√2 is an irrational number

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Updated on: September 25th, 2023 6+√2 is an irrational number. Rational Numbers can be expressed as fractions or decimals, while irrational numbers have non-repeating, non-terminating decimal representations and cannot be expressed as simple fractions. To prove that 6+√2 is an irrational number, we need to signify that it cannot be expressed as a simple fraction or a ratio of two integers.

Table of content ## Difference Between Irrational Number and Rational Number

Rational numbers are numbers that can be expressed as the ratio of two integers (p/q), where q is not equal to zero. These numbers can be written in the form of fractions and can be finite or repeating decimals. For example- 1/2, 3/4, and 0.25 are all rational numbers.

On the other hand, irrational numbers cannot be expressed as a simple fraction or a ratio of two integers. These numbers have non-repeating, non-terminating decimal representations. They are typically represented by square roots (√), cube roots (∛), or other mathematical notations. Examples of irrational numbers include √2, π (pi), and e (Euler’s number).

Solution

Let’s assume that 6 + √2 is a rational number and can be expressed as a fraction a/b,

6 + √2 = a/b

We can rearrange the equation:

√2 = (a/b) – 6

Squaring both sides of the equation to eliminate the square root:

2 = ((a/b) – 6)2

2 = (a2 / b2) – 12(a/b) + 36

Multiplying by b2 to eliminate the denominator:

2b2 = a2 – 12ab + 36b2

a2 = 2b2 + 12ab – 36b2

2c2 = (2d)2 + 12(2d)(2c) – 18(2d)2

2c2 = 4d2 + 96dc – 72d2

c2 = 2d2 + 48dc – 36d2

Therefore 6 + √2 cannot be expressed as a rational number.

Hence it is proved that 6 + √2 is an irrational number.

Summary

## Prove that 6+√2 is an irrational number.

6 + √2 is an irrational number. To prove that 6+√2 is an irrational number, it cannot be represented as the quotient of two integers or in the form of a simple fraction.

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