## The irrational root

There are many proofs that the square root of 2 is irrational. Here are two related ones.

### Latest version

This version simply finesses the "lowest common denominator" presentation.

**Proof**

The square root of 2 is irrational.

*Why?*

Because the square root of 2 is not rational.^{1}

*Why?*

Because assuming the square root of 2 is rational produces a contradiction.

*Why?*

Because if the square root of 2 is rational, then it has the form *a**/b* for positive integers *a* and *b*. If we claim *a* and *b* have no common factors, we discover that it does have common factors.

*Why?*

If (*a**/b*)^{2} = 2 then *a* and *b* are both even, hence have a common factor, contradicting our assumption that they have no common factors.

### Assume ratio has no common factors

This is a standard proof that relies on “proof by contradiction.” It can be found in many places, including [1]. Here I rewrite it in my Q&A fashion.

**Proof**

The square root of 2 is irrational.

*Why?*

Because it's not rational.

*Why?*

Because assuming it's rational produces a contradiction.

*Why?*

Any rational number can be written as a ratio of two integers that have no common factors. But if you assume the square root of 2 is such a ratio, you discover it has a common factor of 2. Contradiction.

*Why?*

Let the square root of 2 be the ratio *p**/q*, with *p* and *q* as co-prime integers (no common factors). Square the root, producing *p*^{2}*/q*^{2} = 2, then multiply both sides by the denominator. This leaves *p*^{2} equalling 2*q*^{2}. Hence *p*^{2} is even. Hence *p* is even. Hence *q* is even.

*Why?*

Since *p*^{2} is even, this must equal (2*r*)^{2} for some integer *r*. That means 4*r*^{2} = 2*q*^{2}. Divide both sides by 2. This means 2*r*^{2} = *q*^{2}. Hence *q*^{2} is even, so *q* is even too.

*Why?*

An even squared integer implies an even integer because of the contrapositive—an odd integer when squared produces an odd square: (2*k*+1)^{2} = (4*k*^{2} + 4*k* + 1) = 2(*k*^{2} + *k*) + 1.^{2}

### Don't assume ratio has no common factors

Roger Penrose (n. 3.2, 68 in [4]) notes that the claim that a ratio can be put in lowest terms, while true, is about as hard to prove as the separate proof he uses to show (51–53) that the square root of 2 is irrational.

This different proof doesn't assume a ratio can be simplified to having no common terms. What it does say is that just as *a* and *b* are solutions to (*a**/b*)^{2} = 2, so are *c* and *d*, *e* and *f*, etc., with *c* smaller than *a*, *e* smaller than *c*, etc., and a similar descending series for the successive denominators.

This process never stops, but these solutions are supposed to be positive integers. You can't start with a positive integer and end up with an infinite number of smaller positive integers.

Hence it's a proof by contradiction.

**Proof**

The square root of 2 is irrational.

*Why?*

Because it's not rational.

*Why?*

Because assuming it's rational produces a contradiction.

*Why?*

Since we're looking for a rational square root of 2, we need two positive integers^{3} such that (*A**/B*)^{2} = 2. If we assume that (*a**/b*)^{2} is one such solution, we can find an endless number of other solutions, each with positive integers. But each new solution has smaller positive integers, but they can't go on forever and stay positive.

*Why?*

We start with *a**/b* as the square root of 2. The square of this root should equal 2: *a*^{2}*/b*^{2} = 2. But we discover that *a* = 2*c* and *b* = 2*d* for some *b*, *d*, which in turn are also even. So *a*, *b*, *c*, *d*,… are all even, and we have two series of endlessly descending positive integers: *a*, *c*, *e*, *g*,…, and *b*, *d*, *f*, *h*,… .

*Why?*

We assume the square root of 2 is *a**/b*, the ratio of two positive integers. Square it and it equals 2: *a*^{2}*/b*^{2} = 2, so *a*^{2} = 2*b*^{2}. But that makes *a* even, so *b* is even, so *c* is even, so *d* is even, and so on: *b*^{2} = 2*c*^{2}, *c*^{2} = 2*d*^{2}, and so on forever, with a > b > c > d > … .

*Why?*

If you square an odd number you get an odd number. Inversely, if you have an even square it must have come from an even integer. So *a*^{2} = 2*b*^{2} implies *a*^{2} is even, hence *a* is even. But that means *a* = 2*c* for some integer *c*. Therefore 4*c*^{2} = 2*b*^{2}. At this point you divide both sides by 2, hence *b*^{2} = 2*c*^{2}. You go through the same process, and when you divide both sides by 2 each variable introduced is smaller than the older ones: *a*^{2} is twice *b*^{2}, which is twice *c*^{2}, which is twice *d*^{2}, etc., etc.

*An Introduction to Algebraic Structures*. New York: Dover, 1989 [orig. 1969].

*Bridge to Abstract Mathematics*. On-line publication, 2003. http://www.science.marshall.edu/lawrence/syllabi/bam.pdf.

*The Road to Reality: A Complete Guide to the Laws of the Universe*. New York: Knopf, 2005 [orig. 2004].

## Advisory

*Please note that I am not a mathematician and so the presentation of proofs that I make may be deeply flawed. I'm using this writing process to figure out what I'm reading. Please consult more authoritative sources as well.*

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