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Proof: sum of rational & irrational is irrational

CCSS.Math:

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 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 we lead to if this leads to any form of contradiction so let's assume that this is going to give us a rational 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 let's call and their sum gives us another rational number well let's give me let's express that as a ratio of two other integers m and n so we're saying that a over B a over B plus X plus X is equal to M over N is equal to M over N well another way of thinking about it we could subtract a over B from both sides and we would get our irrational number X is equal to M over N minus a over B minus a over B which is the same thing as n times B in the denominator n times B and then let's see M over N is the same thing as M B over n B so this would be M B I'm just adding these two fractions M B M B - - let's see a over B the same thing as n times a 2 over N times B so minus n times a minus n times n times a all I did is I added these two fractions I found a common denominator so let me make it clear I multiplied this one B and B and then I multiplied this one N and N and 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 and a is an integer the difference of two integers this whole thing is going to be an integer 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 the ratio of two integers so this tells us that X must be rational 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 just so a rational plus an irrational irrational must be must be irrational