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Studying for a test? Prepare with these 3 lessons on Irrational numbers.
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Proof: sum of rational & irrational is irrational

Video transcript
So I'm curious as to what happens if I were to take a rational number and I were to add it to an irrational number. Is the resulting number going to be rational or irrational? Well, to think about this, let's just assume it's going to be rational and then see if this leads to any form of contradiction. So let's assume that this is going to give us a rational number. So let's say that this first rational number we can represent as the ratio of two integers, a and b. Let's call this irrational number, let's just call this x. And their sum gives us another rational number. Well, let's express that as the ratio of two other integers, m and n. So we're saying that a/b plus x is equal to m/n. Well, another way of thinking about it-- we could subtract a/b from both sides and we would get our irrational number x is equal to m/n minus a/b, which is the same thing as n times b in the denominator. And then let's see. m/n is the same thing as mb over nb. So this would be mb. I'm just adding these two fractions. mb minus-- Let's see. a/b is the same thing as n times a over n times b. So minus n times a. All I did is I added these two fractions. I found a common denominator. So to make it clear, I multiplied this one b and b, and then I multiplied this one n and n then I just added these two things, and I got this expression right over here. So this denominator is clearly an integer. I have the product of two integers. That's going to be an integer. That's going to be an integer. And then this numerator, mb, is an integer. na is an integer. The difference of two integers. This whole thing is going to be an integer. So it looks like, assuming that the sum is rational, that all of a sudden we have this contradiction. We assumed that x is irrational, we're assuming x is irrational, but, all of a sudden, because we made that assumption, we're able to assume that we can represent it as this ratio of two integers. So this tells us that x must be rational. And that is the contradiction. That is a very large contradiction right over there. The assumption was that x is irrational. Now we got that x must be rational. So, therefore, this cannot be the case. A rational plus an irrational must-- so this is not right-- a rational plus an irrational must be irrational. Let me write that down. So a rational plus an irrational must be irrational.