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### Course: Algebra 1>Unit 15

Lesson 3: Proofs concerning irrational numbers

# Proof: there's an irrational number between any two rational numbers

Sal proves that when given any two rational numbers, no matter how close, we can find an irrational number that lies between them. Created by Sal Khan.

## Want to join the conversation?

• why 1/sqrt(2) = sqrt(2)/2 ?
(22 votes)
• I found another way, which uses basic lessons from Khan Academy about roots and fractions.

1/sqrt(2) = 1/sqrt(4/2) = 1/(2/sqrt(2)) = 1 * sqrt(2)/2 = sqrt(2)/2
(8 votes)
• May I add a simple logic to all of this? This way more people will understand. Can you confirm that my logic holds true, Sal?
Here is one way of looking at the claim we are attempting to prove
Question: Can we cover each distance on the number line with the set of rational numbers, or is the set of rational numbers insufficient?
It might originally appear that the set of rational numbers suffices, for, after all, we can always make smaller and smaller rationals. Say we start at 1 unit distance, and we divide this distance in 2 repeatedly, we should get smaller and smaller units, until it becomes infinitely small.
However, by finding that in 1 place on the number line, there exists a irrational between 2 rationals, we know that the set of rationals is insufficient to cover all distance on the number line.
Because there is nothing special about one specific place in the number line over any other place (accomplished technically by adding r1 and multiplying by r2-r1), we know that between all 2 rationals there exists an irrational.
Please confirm this logic. Thanks
(21 votes)
• You've given a reasonable summary of Sal's proof.
(6 votes)
• How to prove rational number+irrational number is an irrational number
(8 votes)
• First prove a rational - rational = rational. A rational is a fraction a/b where a and b are natural numbers. Let a/b and c/d be two rational numbers...
a/b + c/d = (ad + bc)/bd
ad + bc = natural number
bd = natural number
so (ad + bc)/bd is a rational number
So a rational - a rational = rational...REMEMBER THIS
Now let a/b be a rational number and x an irrational number
Suppose a/b + x = c/d where c/d is a rational number
So we have...
a/b + x = c/d
x = c/d - a/b
But you know (from our first proof) that c/d - a/b is a rational number.
So, x is a rational number AND x is an irrational number. Contradiction!

Therefore, our assumption that a rational + an irrational = rational is false.
Therefore, a rational + an irrational = irrational

This method is called reductio ad absurdum or proof by contradiction. Read it up, it's very interesting.
(22 votes)
• Why do I need to prove this? What if I just write a number like 84,5426578422158456622 Is it not an irrational number?
(2 votes)
• ``Good question !First of all, since your number can be written with a finite amount of rational digits it is definitely not irrational. Irrational numbers can not be written with a finite amount of non repeating digits or an infinite amount of repeating digits, i.e. they do not show a pattern when expressed with rational numbersThen to the second point, "Why": Saying things like "What if ..." or "is it not..." is not enough for a mathematical proof. A proof in Math has to be absolutely stable against all reasonable doubt. This comes from a peculiarity of mathematical proofs: They are disproven as soon as you find ONE counterexample. Unlike e.g. in Medicine, where basically all "security" comes from statistics, a mathematical proof is either perfect (=absolutely secure against all counter examples) or not valid at all.``
(23 votes)
• proof important for sat or not?
(6 votes)
• The SAT is mainly short answers, so proofs aren't really needed, usually just the result
(8 votes)
• so if the square root of like 49 is 7 then is it irrational or rational
(7 votes)
• It's rational because after you take the square root of 49 and get 7, the 7 could be written as a ratio, or fraction, equal to 7/1.
(6 votes)
• Is there an irrational number between 0.999999999... and 1? I have thought about this problem for a while but can't find an irrational number between them although Sal says there is. Please help.
(4 votes)
• It's because by simple arithmetic logic, 0.99999..... is equal to 1. Let's see how:
``let x = 0.9999....10x = 9.9999...10x = 9 + 0.9999...10x = 9 + x9x = 9x = 1``

Now whether it's really true or not, beyond my scope. It means that 0.9999... is so, so close to 1, that it equals 1. So there wouldn't be an irrational number between them, as both of them are the same number.

Hope it helps. :)
(3 votes)
• This one is much easier to understand than the last proof for dumb like me :)
(5 votes)
• Can I find a countably infinite number of irrationals between any two rational numbers ?
(4 votes)
• It is somewhat possible. Look at this link if you have trouble understanding. https://www.youtube.com/watch?v=h2YfuPCP3uA
(1 vote)
• x + y / 2 is a rational number between x and y .

Why? Please Explain.
(1 vote)
• This is not true, I can show this by example. Suppose x = 10 and y = 6. So 10 + 6/2 = 13 which is not between 6 fand 10..
However, if you have (x + y)/2 which is probably what you started with, this just gives the average of two numbers, and the average of any two numbers is always 1/2 the way between the two numbers.
(4 votes)

## Video transcript

What I want to do in this video is prove that between any two rational numbers-- so let's say that's a rational number there, and then let's say that this is another rational number that is larger than this one right over here-- that between any two rational numbers, you can find an irrational number. So that number right over there is irrational. You can find at least one irrational number. And that's kind of crazy, because there's a lot of rational numbers. There's an infinite number of rational numbers. So we're saying between any two of those rational numbers, you can always find an irrational number. And we're going to start thinking about it by just thinking about the interval between 0 and 1. So if we think about the interval between 0 and 1, we know that there are irrational numbers there. In fact, one of them that might pop out at you is 1 over the square root of 2, which is the same thing as the square root of 2 over 2, is equal-- I shouldn't say equal, is roughly, is approximately equal to 0.70710678118. And I could just keep going on and on and on and on and on and on. This thing does not repeat. But the important point is, it's clearly between 0 and 1. So I could write 1 over the square root of 2 is clearly between 0 and 1. So the way that I'm going to prove that there's an irrational number between any two rational numbers is I'm going to start with this set of inequalities, and I'm going to manipulate it so I end up with an r1 over here and an r2 over here. And then from 1 over the square root of 2, I would have manipulated this to construct that irrational-- at least one of the irrational numbers that's between those two rational ones. So instead of making this an interval between 0 and 1, let's make this an interval between 0 and the difference between these two numbers. So the distance between r1 and r2 is r2 minus r1. So let's multiply both sides of this-- or all three parts of this inequality, I guess I could say, by r2 times r2 minus r1. So let's do that. So if you multiply this, 0 times r2 minus r1, well you're just still going to have 0 there, is less than-- And we know that r2 is greater than r1, so r2 minus-- let me make it clear what we're doing. We're going to multiply everything times r2 minus r1. r2, we're assuming, is greater than r1, so this thing right over here is going to be greater than 0. So if you multiply the different sides of an inequality by something greater than 0, you don't switch the inequality. So 0 times that is 0, 1 over the square root of 2 times that is going to be 1 over the square root of 2 times r2 minus r1. And then that's going to be less than-- well, 1 times that is just going to be r2 minus r1. And now, we just have to kind of shift everything over. So let's add r1 to all sides of this. So if we add something to all parts of the inequality, then that's also not going to change the inequality. So we're going to add r1 over here. We can add r1 over here. And we can add r1 over there. And so on the left-hand side, we have r1 is less than r1 plus-- let me just copy and paste all of this so I don't have to keep changing colors. Whoops, that's not what I wanted to do. Let me do this. There you go. All right. That should be pretty good. So copy and paste that. r1 plus this, plus that-- let me write the plus down-- plus that, is less than-- that one is a different shade of blue-- is less than-- well, what's r1 plus r2 minus r1? Well, that's just going to be r2. So I've just shown you that you give me any two rational numbers, and I'm assuming r2 is greater than r1, I have just constructed an irrational number that's going to be between those two rational numbers. You take r1, you take the lower of the rational numbers, and to that you add 1 over square root of 2 times the difference between those two rational numbers, and you are going to get this right over here is an irrational number. You're saying hey, how do I know that this thing-- how can I be satisfied that this thing is irrational? Well, we've already seen. You take the product of an irrational and a rational, you get an irrational number. You take the sum of an irrational number and a rational number, you get an irrational number. So we've constructed an irrational number that's between these two rationals.