How do we know √2 is irrational?

[Vort]

The simple and elegant proofs are always the best, and this proof is almost as simple and elegant as they come. About the only thing you need to assume is that Pythagoras' Theorem is true (that is, the square of a right triangle's hypotenuse is equal to the sum of the squares of its legs).

Consider a right triangle with legs of length 1 unit (you can use any unit you want). The square of each leg is 1 square unit, so we know the square of the hypotenuse is 2 square units. Therefore, the length of the hypotenuse is √2 units. Our job is to prove that the length of the hypotenuse is "irrational", which means that it cannot be represented by a simple fraction a/b where a and b are integers and a/b is in "simplest form" -- that is, you can't simplify it any more, so that it cannot be 4/6, but instead would be 2/3.

Before we get to the proof, note that the stipulation of "simplest form" means that a and b cannot both be even. We know this because if they were both even, you could divide each by 2 and simplify the fraction -- but we have stipulated that the fraction is already in simplest form.

(You don't actually have to stipulate simplest form; you just have to recognize that the fraction has to be reducible to simplest form, which is pretty obvious if we're talking about a real, honest-to-goodness fraction of integers. But the proof is more straightforward if you just stipulate simplest form.)

Here's the proof:

Let us assume that we can divide the legs and hypotenuse into some integer number of units such that, if we call the leg length a and the hypotenuse length b, a and b are each integers. This may be a pair of relatively small numbers, or they may be extremely large, much greater than a hundred zillion. But we assume that there are some such numbers.

Now, from Pythagoras' Theorem, we know that a² + a² = b², or 2a² = b². This means that b² must be even, because it's equal to two times a², which is an integer, and two times any integer is even. And since b² is even, b must also be even, because the root of an even square must itself always be even. (You can never get an even number by multiplying an odd number times itself.)

We already decided that the leg length and the hypotenuse length cannot both be even. (Remember: Simplest form.) Therefore, since the hypotenuse length b is even, the leg length a MUST be odd. No other possibility.

Since b is even, we know that it equals 2x, where x = ½b and is itself an integer. So, since 2a² = b², we can substitute b = 2x to get:

2a² = (2x)² = 4x²

This simplifies to

a² = 2x²

where x is an integer. So therefore, we know that a must be even, because it's equal to two times an integer.

But wait! We already decided that b is even, so a CAN'T be even! Yet we just proved that a IS even! So what does that mean?

Simple. It means that our original assumption is false: √2 cannot be represented by a fraction a/b where a and b are both integers and a/b is in simplest form. That is to say, √2 is not rational.

We used simple algebra in the above proof, but note that you can do the same proof without using any algebra. For more information, see the utterly brilliant

. Edited March 29, 2013 by Vort

[Jennarator]

How do we know √2 is irrational?

I has been throwing tantrums, acting up, not listening to reason....

[Just_A_Guy]

How do we know √2 is irrational?

Because its political opinions differ from my own?

[wyarwehere]

To quote Pam...

Math is evil.

[Dravin]

A fantastic example of proof by contradiction.

[skippy740]

[Vort]

π responds: "She's such a square. She thinks she's a 10, but only in her own base."

[JimmiGerman]

The simple and elegant proofs are always the best, and this proof is almost as simple and elegant as they come. About the only thing you need to assume is that Pythagoras' Theorem is true (that is, the square of a right triangle's hypotenuse is equal to the sum of the squares of its legs).

Consider a right triangle with legs of length 1 unit (you can use any unit you want). The square of each leg is 1 square unit, so we know the square of the hypotenuse is 2 square units. Therefore, the length of the hypotenuse is √2 units. Our job is to prove that the length of the hypotenuse is "irrational", which means that it cannot be represented by a simple fraction a/b where a and b are integers and a/b is in "simplest form" -- that is, you can't simplify it any more, so that it cannot be 4/6, but instead would be 2/3.

Before we get to the proof, note that the stipulation of "simplest form" means that a and b cannot both be even. We know this because if they were both even, you could divide each by 2 and simplify the fraction -- but we have stipulated that the fraction is already in simplest form.

(You don't actually have to stipulate simplest form; you just have to recognize that the fraction has to be reducible to simplest form, which is pretty obvious if we're talking about a real, honest-to-goodness fraction of integers. But the proof is more straightforward if you just stipulate simplest form.)

Here's the proof: (...)

Simple. It means that our original assumption is false: √2 cannot be represented by a fraction a/b where a and b are both integers and a/b is in simplest form. That is to say, √2 is not rational.

I'm not sure if you are only good at logical thinking and good at Algebra or if you are a lawyer. In the second case you certainly would be one of the most demanded attorneys in the United States and certainly a millionaire.

If I only had a staff of eggheads simply to response your "proof"... grumble... And don't tell me a sphere has no edges, because it's a fact that you can bang your head against a bowling ball!

Edited March 30, 2013 by JimmiGerman

[mordorbund]

It's a doomed relationship. They're seeking help from a counselor infamous for causing division.

[skippy740]

[JimmiGerman]

The simple and elegant proofs are always the best, and this proof is almost as simple and elegant as they come. About the only thing you need to assume is that Pythagoras' Theorem is true (that is, the square of a right triangle's hypotenuse is equal to the sum of the squares of its legs).

Consider a right triangle with legs of length 1 unit (you can use any unit you want). The square of each leg is 1 square unit, so we know the square of the hypotenuse is 2 square units. Therefore, the length of the hypotenuse is √2 units. Our job is to prove that the length of the hypotenuse is "irrational", which means that it cannot be represented by a simple fraction a/b where a and b are integers and a/b is in "simplest form" -- that is, you can't simplify it any more, so that it cannot be 4/6, but instead would be 2/3.

Before we get to the proof, note that the stipulation of "simplest form" means that a and b cannot both be even. We know this because if they were both even, you could divide each by 2 and simplify the fraction -- but we have stipulated that the fraction is already in simplest form.

(You don't actually have to stipulate simplest form; you just have to recognize that the fraction has to be reducible to simplest form, which is pretty obvious if we're talking about a real, honest-to-goodness fraction of integers. But the proof is more straightforward if you just stipulate simplest form.)

Here's the proof:

Let us assume that we can divide the legs and hypotenuse into some integer number of units such that, if we call the leg length a and the hypotenuse length b, a and b are each integers. This may be a pair of relatively small numbers, or they may be extremely large, much greater than a hundred zillion. But we assume that there are some such numbers.

Now, from Pythagoras' Theorem, we know that a² + a² = b², or 2a² = b². This means that b² must be even, because it's equal to two times a², which is an integer, and two times any integer is even. And since b² is even, b must also be even, because the root of an even square must itself always be even. (You can never get an even number by multiplying an odd number times itself.)

We already decided that the leg length and the hypotenuse length cannot both be even. (Remember: Simplest form.) Therefore, since the hypotenuse length b is even, the leg length a MUST be odd. No other possibility.

Since b is even, we know that it equals 2x, where x = ½b and is itself an integer. So, since 2a² = b², we can substitute b = 2x to get:

2a² = (2x)² = 4x²

This simplifies to

a² = 2x²

where x is an integer. So therefore, we know that a must be even, because it's equal to two times an integer.

But wait! We already decided that b is even, so a CAN'T be even! Yet we just proved that a IS even! So what does that mean?

Simple. It means that our original assumption is false: √2 cannot be represented by a fraction a/b where a and b are both integers and a/b is in simplest form. That is to say, √2 is not rational.

We used simple algebra in the above proof, but note that you can do the same proof without using any algebra. For more information, see the utterly brilliant

.

Okay. Didn't want to shorten all your proof. But deep in my mind I'm feeling there is s.th. wrong. Let's see. One day I'm giving you an anti-proof. I'm Jimmi German. Mr. Robot! Don't iron a shirt if you have a woman doing that. False robot! Wrongly programmed.

And now your answer: "Robot nothing..." But there is undoubtly seen a robot on your avatar. You're a robot and nothing but a robot. Give me a proof you are none!

Edited April 13, 2013 by JimmiGerman