# Euclid Square Root 2 Irrational

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Before proving $\sqrt{2}$ as irrational, first, let us understand what an Irrational number.

## What is an Irrational Number?

For any integers, an irrational number is a number that can not be represented as a fraction, and irrational numbers have decimal expansions that do not terminate.

The best example of an irrational number is Pi (𝝅) which is has a non-terminating number 3.14159265359.

Here we have to prove the irrationality of $\sqrt{2}$. This proof is a classic example of Proof by Contradiction.

In proof by contradiction, at the start of the proof, the opposite is believed to be valid. The assumption is shown not to be valid after rational reasoning at every stage. This is also known as indirect proof and proof by assuming the opposite.

Euclid developed this proof by contradiction and applied for $\sqrt{2}$ to prove as an irrational number

### Euclid Square Root 2 Irrational Proof

According to proof by contradiction given by Euclid, the first step of the proof, we will assume the opposite is true. In the same way here will we assume that $\sqrt{2}$ is equal to some rational number a/b.

$\sqrt{2}$ = $\frac{a}{b}$……………………(1)

Now, we will square on both sides of equation (1),

($\sqrt{2}$)$^{2}$ = ($\frac{a}{b}$)$^{2}$…………………(2)

We will simplify and rewrite the equation (2) as

2b2 = a2……………………….(3)

If we observe, here the value of a2 will be positive because b2 is multiplied by an even number ‘2’. Since a2 is positive, we can conclude that a is also positive. Since ‘a’ is positive, we can write a=2c where c is any whole number. Since ‘a’ is even number 2 multiplied by any whole number will satisfy the definition of even number.

Now let substitute a=2c in equation (3),

2b2 = (2c)2

2b2 = 4c2………………..(4)

Now divide by 2 into both sides of equation (4), we get

b2 = 2c2……………………(5)

Here b2 is multiplied by 2 and c2 which satisfies the definition of even number. Therefore, b2 is also an even number which concludes that ‘b’ is an even number.

So, we have proved ‘a’ and ‘b’ even numbers.

In the next step, let us assume that b=2d in the same of assuming a=2c which satisfies the even number definition.

Now let us substitute a=2c and b=2d in equation (1) where we have assumed

$\sqrt{2}$ = $\frac{a}{b}$

$\sqrt{2}$ = $\frac{2c}{2d}$

$\sqrt{2}$ = $\frac{c}{d}$ ………………(6)

Now we have obtained c/d which is in simpler form compared to p/q. Also from equations (1) and (6)

a/b = c/d ……………….(7)

Here we can further simplify c/d into say e/f by carrying out the same process. Again e/f will be put through the same process and we obtain g/h is simpler.

Rational number definition states that “a number cannot be simplified indefinitely it has to terminate at some point.

So, the basic assumption that $\sqrt{2}$ is a rational number will fail here. So the answer contradicts the basic assumption that $\sqrt{2}$ as a rational number is unreasonable.

So, we can conclude that the contradiction has been reached that $\sqrt{2}$ is not a rational number.

Hence we have proved that $\sqrt{2}$ is irrational.