- Proof: sum & product of two rationals is rational
- Proof: product of rational & irrational is irrational
- Proof: sum of rational & irrational is irrational
- Sums and products of irrational numbers
- Worked example: rational vs. irrational expressions
- Worked example: rational vs. irrational expressions (unknowns)
- Rational vs. irrational expressions
The product of any rational number and any irrational number will always be an irrational number. This allows us to quickly conclude that 3π is irrational. Created by Sal Khan.
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- At1:44it is said that you should multiply both sides by the reciprocal. Reciprocals only exist for rational numbers that are nonzero. What would happen if the rational number happens to be zero?
A very subtle counterexample.
Can anyone see how to reword the statement to make it true?(28 votes)
- There's now a correction at the beginning of this video indicating that it only works for non-zero rationals. (At1:55the division by
awould be invalid if
awere zero.) Zero times an irrational is of course the rational zero again.(9 votes)
- So what is an irrational number times another irrational number?(7 votes)
- A irrational number times another irrational number can be irrational or rational. For example, √2 is irrational. But:
√2 • √2 = 2
Which is rational. Likewise, π and 1/π are both irrational but:
π • (1/π) = 1
Which is rational.
However, an irrational number times another irrational number can also be irrational:
√2 • √3 = √6
Which is irrational.
Comment if you have questions.(28 votes)
- an irratinal number can also be expresed as irrational number/1 . so i am confused. is my question valid? because it becomes a ratio of 2 numbers.(4 votes)
- You do not have to stop there, you could divide an irrational by any whole number, √/2/2 and √3/3 are common ones you will see in Math. However, the division of a irrational by a rational will still result in an irrational number. The question is valid, but the answer is not the one you thought. You can divide an irrational by itself to get a rational number (5π/π) because anything divided by itself (except 0) is 1 including irrational numbers.
The issue is that a rational number is one that can be expressed as the ratio of two integers, and an irrational number is not an integer.(7 votes)
- What about irrational times irrational?(4 votes)
- √2 and √3 are both irrational.
√2•√2=2, which is rational.
√2•√3=√6, which is irrational.
So a product of two irrationals can be either rational or irrational.(7 votes)
- if pi = 22/7 then why pi is considered an irrational numaber?(2 votes)
- 22/7 is a close approximation of pi which can be useful for some calculations, but it does not equal pi.(9 votes)
- 501/502 = 0.99800796812...
Is the quotient rational? Is the fraction 501/502 rational? Individually 501 and 502 are rational, because each can be expressed as a fraction?(2 votes)
- Yes, both because it can be explained as as a quotient and because even though it has infinite digits, they do repeat.
the decimal part: 98007968127490039840637450199203187250996015936254 repeats over and over and over and over, just like 1/3 for example. meanwhile an irrational number like the square root of 2 or pi do not have repeating decimals.(2 votes)
- Is it possible that a so called irrational number is actually rational? For instance, how can we know for certain that the digits after the decimal go on in a random pattern forever? Is there a way to prove that it is infinitely precise? Or is it possible they stop after an obnoxiously large number of digits?(2 votes)
- See, rational numbers are in the form of p/q where p and q are coprimes (i.e.,they are both integers and have no common factors.)
But in previous videos, we have already seen that an irrational number like square root of 2 can't be expressed as p/q.
Hence irrational numbers are not rational. So the digits must go in a random pattern forever, otherwise it would be rational number, which is not the case.
Check the proof that sqrt(2) is irrational video @1:30
The proof goes like this -
assume sqrt(2) is rational
=> sqrt(2) = p/q
=> 2 = (p^2)/(q^2)
=> p^2 = 2*(q^2)
=> p is a multiple of 2.
=> p = 2m , where m is an integer.
=> 2*(q^2) = p^2 = (2m)^2
=> 2*(q^2) = 4*(m^2)
=> q^2 = 2*(m^2)
=> q is a multiple of 2.
But wait, p and q are coprimes, and we just proved that p and q are both divisible by 2.
=> Our assumption is wrong.
=> sqrt(2) is irrational.
Hope it helps.(2 votes)
- Why did we multiply by reciprocal?(2 votes)
- I'm lost, because it assumed that rational times irrational is rational but then towards the end he was saying that the assumption was irrational.(2 votes)
What I want to do with this video is do a quick proof that if we take a rational number, and we multiply it times an irrational number, that this is going to give us an irrational number. And I encourage you to actually pause the video and try to think if you can prove this on your own. And I'll give you a hint. You can prove it by a proof through contradiction. Assume that a rational times an irrational gets you a rational number, and then see by manipulating it, whether you can establish that all of a sudden this irrational number must somehow be rational. So I'm assuming you've given a go at it. So let's think about it a little bit. I said we will do it through a proof by contradiction. So let's just assume that a rational times an irrational gives us a rational number. So let's say that this-- to represent this rational right over here, let's represent it as the ratio of two integers, a over b. And then this irrational number, I'll just call that x. So we're saying a/b times x can get us some rational number. So let's call that m/n. Let's call this equaling m/n. So I'm assuming that a rational number, which can be expressed as the ratio of two integers, times an irrational number can get me another rational number. So let's see if we can set up some form of contradiction here. Let's solve for the irrational number. The best way to solve is to multiply both sides times the reciprocal of this number right over here. So this, let's multiply times b/a, times b/a. And what are we left with? We get our irrational number x being equal to m times b. Or we could just write that as mb/na. So why is this interesting? Well, m is an integer, b is an integer, so this whole numerator is an integer. And then this whole denominator is some integer. So right over here, I have a ratio of two integers. So I've just expressed what we assumed to be an irrational number, I've just it expressed it as the ratio of two integers. So now we have x must be rational. And that is our contradiction, because we assumed that x is irrational. And so therefore, since this assumption leads to this contradiction right over here, this assumption must be false. It must be that a rational times an irrational is irrational.