Sal proves that the square root of 2 is an irrational number, i.e. it cannot be given as the ratio of two integers. Created by Sal Khan.
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- How can a fact that the assumed numbers are reducible, proves that root 2 is irrational. I didn't get the logic.(21 votes)
- Games have rules and math has rules, so in that sense, math is like a game. If you follow the rules of math, you win. What do you win? You win a truth, which up till now, you have been calling “a correct answer” – but there is more. . .
You already know many math rules, like keeping an equation balanced, or a negative number multiplied by another negative number results in a positive number etc. etc. . . . . There are lots of these kinds of procedural rules for playing with numbers.
But there are other types of rules that will become more and more important as you get further into math and they have to do with the properties of numbers and consequences of these properties. The most amazing thing about these rules is that they, combined with other rules can lead you to a truth than no one has discovered yet, in other words, this rule based game of exploration can take you where no one has gone before – and it is that aspect which keeps us pushing the boundaries of mathematics. Many new discoveries are beautiful (fractal geometry), or useful (chaos theory), or weird (quantum mechanics) and some are even a bit disturbing (the n-body problem).
Now, the path that leads to a truth in mathematics is called a proof. Guess what? You have been doing proofs all this time, right since you first started to add numbers up until now. For example, if you have a problem like 7x - 10 = 5x + 6, you can prove, using the rules of the game of math, that x can only be equal to 8 in order that 7x - 10 = 5x + 6 becomes a true statement. This is an example of a proof using just the procedural rules. If at each step of the way, you obey the rules, you prove (arrive at the truth) that x = 8.
The proof you are asking about in this video is a proof that uses some properties of numbers and some concepts and their consequences. This takes a bit more considered thinking. The proof that √2 is irrational is the most common introduction to this type of thinking.
So, here we go . . . . .
First, would you agree that any rational number whose numerator and denominator are not co-prime, can be reduced to a co-prime form? (if you don’t agree, look into it, because it is true). Co-prime is just a fancy pants way of saying that the greatest common divisor between two numbers is 1. Lots of times you will hear it said as “reduced form”. So if you have 4/6, you can take out the 2 which is a common term between 4 and 6, to get 2/3 and then the only factor that 2 and 3 have in common is 1, right? This process is true for any rational number, that is, a rational number that is not already in co-prime form can be reduced to co-prime form. So, for example 6/9 = 4/6 = 2/3, the ratio is the same, the result is the same even though the numbers are different. This little detail becomes important in the proof . . . .
A common method of proof is called “proof by contradiction” or formally “reductio ad absurdum” (reduced to absurdity). How this type of proof works is: suppose we want to prove that something is true, let’s call that something S. If we start the proof by assuming that S is false, and then through a series of mathematically sound arguments, we can show that we get a nonsense or contradictory result, well then, that means that the assumption we made that S was false can’t be correct, so S must be true.
In this proof we want to show that √2 is irrational so we assume the opposite, that it is rational, which means we can write √2 = a/b. Now we know from the discussion above that any rational number that is not in co-prime form can be reduced to co-prime form, right? So for the sake of argument, let’s assume that we have done any needed reduction and now a and b are reduced (co-prime), meaning that the only common divisor between the numerator and denominator is 1. Now we square both sides of √2 = a/b to get 2=a²/b² and from there it is a short journey to show that since 2=a²/b², it means that both a and b must be even numbers which means they have a 2 in common. Oops! We said that they only had a 1 in common.
Here is the subtle part:
Even though it is true that having the number 2 in common is in contradiction to the statement that a and b are in reduced form (co-prime), this contradiction is not the big deal. The big deal is that we got to this stage by assuming that √2 was rational in the first place! By assuming that √2 is rational, we were led, by ever so correct logic, to this contradiction. So, it was the assumption that √2 was a rational number that got us into trouble, so that assumption must be incorrect, which means that √2 must be irrational.
Here is a link to some other proofs by contradiction:
as an aside, you might want to look up the definition of mathematical soundness:
- has anybody looking up the digits of sqrt 2 noticed that there is a bit of pi in the sequence of its digits? first you have 1.41 as the first 3 digits of sqrt 2. They are also the first 3 decimal places in pi. Later on in the sequence you have 3141 which are the first 4 digits of pi. If it continues like this do you think that very deep in the sequence of the digits of sqrt 2 that there might actually be the number pi? if so why isn't pi constructable like sqrt 2 is? and why isn't it algebraic like sqrt 3?
constructable number: number you can construct with a straightedge and compass like the Greeks did
algebraic number: number that is a root of a non-zero polynomial in one variable with rational coefficients (or equivalently—by clearing denominators—with integer coefficients)
transcendental: neither algebraic or constructable(10 votes)
- Because the digits of √2 do not repeat forever, it is likely that any desired sequence of digits will eventually occur somewhere in √2. You can likely find the first X digits of pi somewhere in √2. You can likely find your phone number, or any long strings of consecutive zeros. I only say likely though, because it hasn't been proven that irrational numbers behave this way. It goes way beyond the scope of an answer here, but see these articles for some relevant theories: http://www.lbl.gov/Science-Articles/Archive/pi-random.html http://mathworld.wolfram.com/BBPFormula.html(3 votes)
- I get lost at5:34when Sal says 2b^2 = 4k^2. It seems to come out of nowhere. Why does the "2k^2" change to "4k^2?"(5 votes)
- He shows that a is even. So therefore a = 2k, and a^2 = (2k)^2. So:
(2k)^2 = (2k)*(2k) = 2*k*2*k = 2*2*k*k = 4(k^2) = 4k^2(8 votes)
- but odd x odd = even? -3 x -3 = 9(0 votes)
- Can't I write √2 as √2/1?(6 votes)
- Yes but the rule to determine rational numbers is that it can be expressed as a fraction of two integers, and √2 is definitely not an integer.(6 votes)
- Why do we automatically assume that neither a nor b have common factors? Or that a/b is irreducible? I understand that needed to be true to achieve the contradiction but it seems arbitrary that we just decided they would have no common factors and be irreducible.(5 votes)
- All the assumption is saying is that the fraction is fully reduced (so in its simplest form). That is not an unreasonable assumption. Any fraction where a/b did share common factors can be simplified into its simplest form.(6 votes)
- How would we prove that the cube root of 4 is irrational using this method?(3 votes)
- Good question! We can do a similar proof.
Assume that cuberoot(4) is rational. Then we can write cuberoot(4) in the form a/b, where a and b are integers, b is nonzero, and a/b is in reduced form (irreducible).
Cubing both sides gives 4=a^3/b^3 which would give 4b^3=a^3. Since 4b^3 is clearly even, a^3 is even. Since the cube of an odd number is odd, a must be even.
So we can write a=2k for some integer k. Therefore, 4b^3=(2k)^3=8k^3 which gives b^3=2k^3. Since clearly 2k^3 is even, b^3 is even, which then implies b is even (again since the cube of an odd number is odd).
So a and b are even, contradicting the irreducibility of a/b. This proves that cuberoot(4) is irrational.(8 votes)
- Not clear first we assumed a/b is irreducible then, Sal said a/b is even why?(3 votes)
- 1. The assumption that a/b is irreducible simply means that the fraction representing the rational number is in simplest terms.
2. Sal then uses the expression 2b^2 = a^2 to show that a must be even.
( 2b^2 is an even number because it has a factor of 2.
So a^2 is also even because it equals 2b^2.
But a^2 is the same as a*a ---- and only even numbers can be multiplied by themselves to produce another even number.
Therefore, a is an even number.)
3. Sal then uses the expression 2b^2 = (2k)^2 in a similar way to show that b must be even.
4. Because both a and b are even, that means they both have a common factor of 2. (Note that is not the fraction a/b that is even, it is the individual numbers a and b.)
5. But, based on the initial assumption that a/b was irreducible, it is impossible for a and b to have a common factor.
6. Therefore, the assumption that sqrt2 was rational is incorrect. It is therefore irrational.
I hope this answers your question.(8 votes)
- I don't understand this proof concerning irrational numbers. can some please explain it in a simpler way.
- A rational number can be expressed as a ratio between two integers (aka a fraction).
Sal started by assuming sqrt(2) is rational. He then showed that you can’t represent sqrt(2) as a ratio between two co-prime integers, which is a direct contradiction of the definition for rational. Because of this contradiction, sqrt(2) actually can’t be rational and must be irrational instead.(4 votes)
- Is a square root of any prime number always irrational?(2 votes)
- Yes. Think about the definition of prime number. It doesn't have any factors other than itself and 1. So there are no repeating numbers in its factorization thus its square root must be irrational(7 votes)
What I want to do in this video is prove to you that the square root of 2 is irrational. And I'm going to do this through a proof by contradiction. And the proof by contradiction is set up by assuming the opposite. So this is our goal, but for the sake of our proof, let's assume the opposite. Let's assume that square root of 2 is rational. And then we'll see if we lead to a contradiction, that this actually cannot be the case. And if it cannot be the case that is rational, if we get to a contradiction by assuming the square root of 2 is rational, then we have to deduce that the square root of 2 must be irrational. So let's assume the opposite. Square root of 2 is rational. Well, if the square root of 2 is rational, that means that we can write the square root of 2 as the ratio of two integers, a and b. And we can also assume that these have no factors in common. Let's say that they did have some factors in common. If we divided the numerator and the denominator by those same factors, then you're getting into a situation where they have no factors in common. Or another way of saying is that a and b are co-prime. Or another way of saying it is we could write this as a ratio of two integers where this is irreducible, where these no longer share any factors. If you can write anything as the ratio of two integers, then you could obviously simplify it further, factor out any common factors to get it to a point where it is irreducible. So I'm going to assume that my a and b, that this fraction right over here, is irreducible. And this is important for setting up our contradiction. So I'm going to assume that this right over here is irreducible. a and b have no factors in common. Let me write that down because that's so important for this proof. a and-- want to do that same color-- a and b have no factors in common, other than, I guess, 1. So this is irreducible. These two numbers are co-prime. So what does that do for us? Well, let's just try to manipulate this a little bit. Let's square both sides of this equation. So if you square the principal root of 2, you're going to get 2. And that's going to be equal to a squared over b squared. And that just comes from a over b squared is the same thing as a squared over b squared. And now we can multiply both sides of this by b squared. And so we get 2 times b squared is equal to a squared. Now, what does this tell us about a squared? Well, a squared is some number, b squared times 2. So anything times 2 is going-- this is going to be an integer. We assumed b is an integer, so b squared must be an integer, and so you have an integer times 2. Well, that must give you an even number. That must give you an even integer. So this right over here, a squared, must be-- so this tells us that a squared must be even. Now, why is that interesting? Well, a squared is the product of two numbers or is the product of the same number. It's a times a. So this is another way of saying that a times a is even. So what does that tell us about a? Let's just remind ourselves. a is either going to be-- we're assuming a is an integer-- a is either going to be even or odd. We just have to remind ourselves if we multiply an even times an even, we get an even number. If we multiply an odd times an odd, we get an odd number. So we have a number times itself. We got an even number. Well, the only way to get that is if that number is even. So this tells us that a is even. And another way of saying that a is even is to say that a can be represented as the product of 2 times some integer. So let's say some integer k. So where is all of this going? Well, as you'll see, we can then use this to show that b must also be even. So let's think about that a little bit. So let's go back to this step right over here. If we say that a can be represented as two times the product of some integer, and that comes out of the fact that a is even. Then we can rewrite this expression right over here as 2-- I'll do it over here-- 2 times b squared is equal to 2k squared. Instead of a squared, I could write 2k squared. We're claiming, or we're deducing, that, assuming everything we've just assumed, that a is even. So if a is even, it can be represented as a product of 2 and some integer. And then we can write that 2 times b squared is equal to 4k squared. And then you divide both sides by 2. You get b squared is equal to 2k squared. And this tells us that, well, k squared is going to be an integer. You take any integer times 2 you're going to get an even value. So this tells us that b squared is even. So that tells us that b squared is even. Well, if b squared is even, by the same logic we just used, that tells us that b is even. So here's our contradiction. We assumed, in the beginning, that a and b have no common factors other than 1. We assumed that this fraction right over here, a/b, is irreducible. But from that and the fact that a/b must be equal to the square root of 2, we were able to deduce that a is even and b is even. Well, if a is even and b is even, and they both have 2 as a factor, and then this isn't irreducible. You could divide the numerator and the denominator by 2. a and b have a common factor of 2. So let me write this down. So this is just to make it clear. So from this and this, we have a and b have common factor of 2, which means that a over b is reducible. And so that's the contradiction. So you assume that square root of 2 can be represented as an irreducible fraction a/b, irreducible because you can say ratio of two integers right over here, that leads you to the contradiction that, no, it actually can be reducible. So, therefore, you cannot make this assumption. It leads to a contradiction. Square root of 2 must be irrational.