Prove That the Square Root of Two Is Irrational
Rational numbers are numbers that can be expressed as a fraction of two whole numbers, a ratio. An irrational number is a number that does not have this property, it cannot be expressed as a fraction of two numbers. Some of the most famous numbers are irrational - think about <math>\pi</math>, <math>e</math> or <math>\phi</math>. <math>\sqrt{2}</math> is an irrational number, and this can be proven algebraically in a very elegant manner.
Steps
- Assume that<math>\sqrt{2}</math> is rational. Then it can be expressed as a fraction <math>\frac{a}{b}</math>, where <math>a</math> and <math>b</math> are both whole numbers, and <math>b</math> is not <math>0</math>. Furthermore, this fraction is written in simplest terms, meaning that either <math>a</math> or <math>b</math>, or both are odd whole numbers.
- <math>\sqrt{2}=\frac{a}{b}</math>
- Square both sides.
- <math>2=\frac{a^2}{b^2}</math>
- Multiply both sides by <math>b^2</math>.
- <math>2b^2=a^2</math>
- Note that <math>a^2</math> is an even number. <math>a^2</math> is even number because it is equal to two times a whole number. Since <math>a^2</math> is even, <math>a</math> must be even too, because if it were odd, <math>a^2</math> would be odd as well (an odd number times and odd number is always an odd number). <math>a</math> is even, so that means it can be written as two times a certain whole number, or in other words, <math>a=2k</math>, where <math>k</math> is this whole number.
- Substitute <math>a=2k</math> into the original equation.
- <math>2=\frac{(2k)^2}{b^2}</math>.
- Expand <math>(2k)^2</math>. <math>(2k)^2=2^2k^2=4k^2</math>.
- <math>2=\frac{4k^2}{b^2}</math>
- Multiply both sides by <math>b^2</math>.
- <math>2b^2=4k^2</math>.
- Divide both sides by two.
- <math>b^2=2k^2</math>
- Note that <math>b^2</math> is an even number. <math>b^2</math> is even number because it is equal to two times a whole number. Since <math>b^2</math> is even, <math>b</math> must be even too, because if it were odd, <math>b^2</math> would be odd as well (an odd number times and odd number is always an odd number).
- Recognise that this is a contradiction. You have just proven that <math>b</math> is even. However, you have also proven that <math>a</math> is an even number. This is a contradiction because in the beginning of this proof, it was assumed that <math>\frac{a}{b}</math> was written in simplest terms, but if both <math>a</math> and <math>b</math> are even, the numerator en denominator can be divided by 2, which means it was not written in simplest terms. Since this is a contradiction, the original assumption that <math>\sqrt{2}</math> is rational is false, thus leading to the conclusion that <math>\sqrt{2}</math> is irrational.
Tips
- This type of proof is reductio ad absurdum (reduction to absurdity). It attempts to disprove a claim by showing that if the claim were true, it would lead to an absurd, impossible or impractical conclusion.
