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Current time:0:00Total duration:3:15

Proof: product of rational & irrational is irrational


Video transcript

what I want to do in this video is do a quick proof that the that if we take a rational number and we multiply it times an irrational number times an irrational number that this is going to give us that that's going to give us an irrational number in 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 irrational times an irrational get to a rational number and then see by manipulating it whether you can establish that all of a sudden this arash 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 proof by contradiction so let's just assume let's assume assume that a rational a rational x and irrational times an irrational gives us a rational number gives us a rational number so let's say that this so lets to represent this rational right over here let's represent it is the ratio of two integers a over B and then this irrational number I'll just call that X so we're saying a over B times X can get us some rational numbers so let's call that M over N let's call this equaling M over N so I'm assuming that a rational a rational number which can be expressed as a ratio of two integers times an irrational number can get me another rational number so let's see if we can set up a 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 over a times B over a times B over a and what are we left with we get our irrational number X we get our irrational number X being equal to M times B or we could just write that as mb mb over n a over n over n a 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 I've just 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 expressed it as the ratio of two integers so now we have X 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 leads to this contradiction right over here leads to this contradiction we cannot this assumption must be false it must be that a rational times an irrational is irrational a rational times an irrational is irrational