How is it proven that √2 or π are irrational? couldnt they just start repeating a zero after the quintillionth digit forever? or maybe repeat the whole number sequence again after quintillion digits

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im just wondering since irrational numbers supposedly dont end and dont repeat either, why is it not a possibility that after a huge bunch of numbers they all start over again or are only a single repeating digit.

In: Mathematics

11 Answers

Anonymous 0 Comments

It depends on the number. For sqrt(2) a proof by contradiction works.

Assume sqrt(2) is rational and thus sqrt(2) = n/m and assume that n/m are the reduced form (i.e., you can’t simplify the fraction more).

2 = n^2 / m^2

2 m^2 = n^2

Since n^2 is 2 times another number, we know n^2 (and thus n) is even.

Let’s replace n with 2p, which we know is possible since it’s even

2 m^2 = (2p)^2

m^2 = 2p^2

Since m^2 is 2 times another number, we know m^2 (and thus m) is even.

Two even numbers divided by one another cannot be the reduced form of a fraction (since you can divide the numerator and denominator by 2).

This means that there can be no reduced form fraction representing sqrt(2).

Anonymous 0 Comments

The proof that sqrt(2) is irrational is fairly simple.

You assume that sqrt(2) is rational, and is represented by some reduced fraction a/b.

sqrt(2) = a/b
2 = a^2 / b^2
a^2 = 2 * b^2

Since *a*^2 is 2 * *b*^(2), we can infer that *a*^2 is even, and therefore *a* is even. Let’s replace *a* with 2 * *x*.

(2*x)^2 = 2 * b^2
4 * x^2 = 2 * b^2
2 * x^2 = b^2

Since b^2 is 2*x^(2), we can now assume that b^2 is even, and therefore b is even.

We made the assumption at the start that a/b was the simplest form of sqrt(2), but now we know that both A and B are even, which means it is not the most reduced form of the fraction. Thus, our assumption was incorrect, and sqrt(2) cannot be expressed as a fraction, and is therefore irrational.

As for Pi, that’s a much longer proof. It was only proven to be irrational in 1761. You can look at the Wikipedia page to see how complex these proofs are in comparison to sqrt(2).

https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational

Anonymous 0 Comments

A rational number can by definition be expressed as the fraction of two integers p/q where q is not zero. Any decimal section that terminates or starts to repeat will be possible to express with p and q as integers. Proof that a number is irrational tends to be to show it is impossible to express at the fraction of two integers

There are many proofs pi is irrational none of them are simple enough to write here on Reddit but look at [https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational) with 6 different proofs.

There are proof for the square root of 2 too [https://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational) you can show that any square root that is not an integer is irrational

Anonymous 0 Comments

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Anonymous 0 Comments

Related question, and perhaps one I should post as an actual question, but its always confused me how π is “never ending”, but 2πr gives us the circumference.

But if π is never ending, then the resulting circumference value should also be never ending. Or an approximation.

But you can measure the circumference for a given radius manually and arrive at an exact figure.

What am I missing?

Anonymous 0 Comments

ViHart on YouTube has a… Good explanation. I think it would be great if they just slowed down, but if you watch it on half speed, it’s better. The idea is to first assume that it is rational. If it is, it can be written as a/b where a and b are both whole numbers. In fact, all rational numbers have a simplest form. In the simplest form, they share no factors. Example 6/8 is not simplest form because both 6 and 8 are divisible by 2. Therefore 3/4 is the simplest form.

Basically, you can prove that a and b are even (if you start with the assumption that it’s rational). That’s a problem because a and b are the generalized form. Meaning you essentially proven that there is no simplest form. If you were to write √2 as a/b, a and b would have to always be even, forever, no matter how many times you divide by 2. It’s a paradox. When you encounter a paradox, you look at your assumptions. Since we only had one assumption, the work is quite easy from here. Our assumption must have been wrong.

Anonymous 0 Comments

I can’t explain it better than Taylor Swift does:

[https://www.youtube.com/watch?v=meudJkjKEe8](https://www.youtube.com/watch?v=meudJkjKEe8)

Anonymous 0 Comments

What’s not been answered yet is the second part of the question.

If either number started repeating a zero after the quintillionth digit, then it would be expressible as a rational number.

3.1400000… is exactly 314/100, and for any such number written in our number system that stops after a certain number of digits, you can do this.

But we have already proven that neither number can be expressed as a rational fraction, and so we also know that pi and root(2) don’t start repeating a zero after the quintillionth digit.

Anonymous 0 Comments

There are a few different methods of proofs in Mathematics. The one people are referencing in the first couple of responses (for proving √2 is irrational) is a Proof by Contradiction. For that proof you start with an assumption (often the opposite of what you are trying to prove). Then you go through steps until you reach a contradiction. In the case of the √2 proof, it is that your reduced fraction of a/b is not reduced.

Once you get the contradiction, your assumption must be wrong. And since your assumption is binary (it is only one of two options), the opposite of your assumption must be correct.

The same type of proof is used to prove there are an infinite number of prime numbers.

Anonymous 0 Comments

Because if a decimal number starts repeating at some point, then it means that it could also be expressed as some fraction. It may be an enormous fraction, two numbers with hundreds or thousands of digits, but it would be a fraction nonetheless. And as the other posts show, we can prove that √2 cannot be represented by any fraction.