#UnderstandingIrrationalNumbers #ProvingIrrationality #π #√2 #Mathematics
Have you ever wondered how it is proven that √2 or π are irrational numbers? 🤔 It’s a common question that many people have when learning about these fascinating mathematical concepts. The idea that these numbers go on forever without repeating can be a bit mind-boggling, and it can be tempting to think that maybe they start repeating after a certain point. But in reality, there are incredibly deep and intricate proofs that show why these numbers are indeed irrational.
Let’s dive into the fascinating world of irrational numbers and explore how mathematicians have proven that √2 and π are irrational.
##Defining Irrational Numbers
To understand why √2 and π are irrational, we first need to understand what it means for a number to be irrational.
– **Irrational numbers** are real numbers that cannot be expressed as a ratio of two integers. In other words, they cannot be written as a fraction.
– They go on forever without repeating, and they cannot be represented by a finite or repeating decimal.
##Proving the Irrationality of √2
The proof that √2 is irrational dates back to ancient Greece and is a classic example of a proof by contradiction.
1. Assume the opposite: Let’s assume that √2 is rational and can be written as a fraction a/b, where a and b are integers with no common factors.
2. Square both sides: If we square both sides of the equation √2 = a/b, we get 2 = a^2/b^2.
3. Simplifying: Rearranging the equation gives us a^2 = 2b^2. This means that a^2 is even, so a must also be even. Let’s say a=2k, where k is also an integer.
4. Substituting: Substituting a=2k back into the equation gives us (2k)^2 = 2b^2, or 4k^2 = 2b^2. Dividing both sides by 2 gives us 2k^2 = b^2, which means b^2 is also even, so b is also even.
Now, we have reached a contradiction because we previously assumed that a and b have no common factors, yet we have just shown that they are both even. This contradiction proves that our initial assumption – that √2 is rational – must be false, meaning that √2 is indeed irrational.
##Proving the Irrationality of π
Proving the irrationality of π is an incredibly complex and fascinating endeavor. While a simple proof like the one for √2 is not available for π, mathematicians have established its irrationality using highly advanced techniques.
– **Lindemann–Weierstrass theorem**: This theorem states that if α is a non-zero algebraic number, then e^α is transcendental. Since π is the solution to the algebraic equation sin(π) = 0, it is considered transcendental and, therefore, irrational.
– **Infinite series and limits**: Another approach to proving the irrationality of π involves using infinite series and calculus to show that no rational number can accurately represent the value of π.
These complex mathematical techniques provide rigorous and compelling evidence that π is indeed irrational.
##Why Irrational Numbers Don’t Repeat
The reason why irrational numbers like √2 and π don’t start repeating or have a pattern emerge after a certain point lies in their inherently non-repeating and non-terminating nature. While it can be difficult to wrap our heads around this concept, the proofs for their irrationality demonstrate why this is the case.
– **Non-repeating nature**: Irrational numbers cannot be expressed with a finite or repeating pattern, which sets them apart from rational numbers that can be written as a fraction.
– **Infinite decimal expansion**: The infinite, non-repeating decimal expansion of irrational numbers is a key characteristic that distinguishes them from rational numbers.
##Conclusion
In conclusion, the proofs for the irrationality of √2 and π are deeply rooted in advanced mathematical concepts and techniques. These proofs make use of logical reasoning, algebra, calculus, and transcendental numbers to establish the irrationality of these fascinating numbers.
While it may seem counterintuitive at first, the evidence for their irrationality is robust and well-established in the field of mathematics. The non-repeating and non-terminating nature of irrational numbers sets them apart from their rational counterparts, making them a captivating and fundamental aspect of mathematics.
`0.333333…` is a never-ending number, but it is **rational** because it can be written as a *ratio* of 1:3. One-third.
`0.428571428571…` is 3:7. Three-sevenths.
ANY number that repeats after the decimal point is a rational number that can be written as the ratio of two integers `a/b`.
An **irrational** number cannot be written as a ratio, BUT some irrational numbers can still be *exactly* represented as the solution to an equation, called *algebraic* numbers.
The number `√2`is *exactly* the solution to the equation `x² = 2`. But we know the decimal expansion never repeats, so there are no integers `a` and `b` that can represent it *exactly.*
We can easily prove an algebraic number to be irrational by contradiction; we assume that `√2` IS rational, and we end up with a logical contradiction as explained in a different comment.
The proof that `π` is irrational is significantly harder because `π` is a **transcendental** number – it cannot be written as the solution to any simple equation; it *transcends* algebra.
All transcendental numbers (`π` and `e` are the most well known) are irrational by their very nature – they just … exist as constants, (meta)physical properties of the universe.
It’s a source of philosophical angst, too – what if `π` was *not* `3.14159 …` but instead exactly `3`? Or anything else? What would the Universe look like?
If you really feel like doing your brain in, this is a bit beyond 5-year-old understanding, but [Matt Parker talks about all the different numbers.](https://www.youtube.com/watch?v=5TkIe60y2GI)
EDIT: typos
Answer: Well first off this is EIL5, not repeat misunderstood textbook content. So let’s try to do that.
An irrational number is a value that cannot be made using two values. Like 1+1+1=3 doesn’t follow that rule, however 1+2=3 does so 3 is rational.
So why isn’t PI rational? Because you said PI, not 3.14 which is a rational value, eg 3+0.14=3.14. Like 3.14159 is rationall and we can prove it using something like 2.14+1.00159. But PI isn’t really a *number*, it’s a calculation. So while any given numerical total of PI is a rational value, PI itself is not.
So what **is** PI? Go ahead and try to measure a circle with a string. You’ll find out you can’t quite get the ends to line up. And how do you measure the string, it’s length because the inside of it is shorter. So to accurately measure a circle someone came up with the idea to take four sticks and make a square and you know everything about the size of that box.
Then you break those lines in half, creating a zizag pattern that looks a little more like a circle. Since you know the size of those lines, you can still calculate everything else about the shape. Break those lines in half a few more times and the human eye sees a circle that’s still mathematically just a bunch of easy to calculate lines.
And there lies the problem, since each calculation is based off the last. **PI** itself cannot be rational. And since numbers can be infinitely halved, calculating PI also has no end. Which is also why PI is infinite, even if 3.14 is not.
The other comments don’t seem to address the root of your question. You’ve got the logic backwards. Numbers aren’t irrational because their decimal representation doesn’t repeat. Their representations don’t repeat because they are irrational. A number is irrational if it isn’t rational, and a number is rational if it can be represented as a fraction of integers. The infinite and nonrepeating decimal expansion is a consequence that can be delivered from that definition.
We don’t determine if a number is irrational by calculating lots of digits and seeing if they repeat. We use other properties of the number to demonstrate that it is irrational. The numerous other comments showing how to do so for the square root of 2 demonstrate how it’s done.
I ran across this fun [visualization](https://www.youtube.com/shorts/aUDYWYqtAR4) a while back. It pretty much shows how an irrational number doesn’t “rationalize”.
> How is it proven?
Mathematicians assumed that √2 is rational, uses logical steps to arrive at a contradiction, and concluded that √2 is not rational.
Why does it work?
Under our set of logic, if you assume X and reach an obviously false result (contradiction), then you can conclude that X is wrong.
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.
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.
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.
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.
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?
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).
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