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### Sqrt2 Irrational

```
Date: 05/17/99 at 12:53:00
From: G.Arjun
Subject: Real Numbers

I would like to know the answer to the question, How do you prove that
the square root of 2 (or other roots of other real numbers) is
irrational? I am not able to understand the argument.
```

```
Date: 05/18/99 at 15:13:35
From: Doctor Floor
Subject: Re: Real Numbers

Hello, G. Arjun,

Thanks for your question!

To prove that sqrt(2) [sqrt means square root] is an irrational number,
we have to show that it cannot be written as a/b, where a and b are
integers.

Such a proof has to be done indirectly:

1. We assume that sqrt(2) could be written as a/b for integer a and b;
2. We show that this leads to a contradiction. So something impossible
has to be derived from our assumption.

The fact that we derive someting impossible from the fact that
sqrt(2) = a/b shows that sqrt(2)=a/b must be false, and we have
proven the theorem.

Let's assume that sqrt(2) could be written as a/b for integer a and b.
We can derive:

sqrt(2) = a/b [sqare both sides]
2 = a^2/b^2 [multiply by b^2; ^2 means squared]
2*b^2 = a^2
a^2 = 2*b^2

Now we have two integers in the equation, 2*b^2 and a^2. We can factor
these integers into primes, and find:

a^2   = 2^m * {product of odd primes}
2*b^2 = 2^n * {product of odd primes}

Because a^2 is a square, m has to be even.
Because b^2 is a square, and "2*" brings an additional factor 2,
n has to be odd.

So m and n are not equal.

But if a^2 = 2*b^2, then the factorisation into primes of these two
has to be equal. So, m and n should be equal.

We have found a contradiction. And our assumption, that sqrt(2) = a/b
for integers a and b, cannot hold. This proves that sqrt(2) is
irrational.

I hope this helps you to understand.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/17/99 at 14:42:20
From: Wayne
Subject: Square root of two

Can you prove to me how the square root of two is irrational?
```

```
Date: 05/17/99 at 16:20:17
From: Doctor Rob
Subject: Re: Square root of two

There are a couple of ways to do that. Both are proofs by
contradiction; that is, they assume that sqrt(2) is rational, and
derive an impossible conclusion from that assumption. Here is one:

Assume sqrt(2) = a/b, reduced to lowest terms (that is, a and b have
no common factor besides 1). Then 2 = (a/b)^2 = a^2/b^2, so
2*b^2 = a^2. That means that a^2 is even, which implies that a is
even. Then write a = 2*c, a^2 = 4*c^2 = 2*b^2, so 2*c^2 = b^2. That
means that b^2 is even, which implies that b is even. This means
that a and b have the common factor 2, which is a contradiction.
Thus no such fraction a/b can exist, and sqrt(2) is irrational.

Here's another:

Assume sqrt(2) = a/b, with a and b positive integers, and least
denominator among all such fractions. Then

2*b^2 = a^2,
2*b^2 - a*b = a^2 - a*b,            4*b^2 > 2*b^2 = a^2 > b^2,
b*(2*b-a) = a*(a-b),                2*b > a > b,
(2*b-a)/(a-b) = a/b = sqrt(2),      b > a - b > 0  and  2*b - a > 0.

Now (2*b-a)/(a-b) is a fraction equal to sqrt(2) with positive integer
numerator and denominator, but smaller denominator than a/b has.
This is a contradiction. Thus no such fraction a/b can exist, and
sqrt(2) is irrational.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents

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