If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Proof: product of rational & irrational is irrational

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.

## Want to join the conversation?

• At it 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?
• There's now a correction at the beginning of this video indicating that it only works for non-zero rationals. (At the division by `a` would be invalid if `a` were zero.) Zero times an irrational is of course the rational zero again.
• So what is an irrational number times another irrational number?
• 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.
• 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.
• 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.
• What about irrational times irrational?
• √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.
• if pi = 22/7 then why pi is considered an irrational numaber?
• 22/7 is a close approximation of pi which can be useful for some calculations, but it does not equal pi.
• Couldn't m/n divided by a/b equal a rational number, x?
• 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?
• 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.
• 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?
• 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 @

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.