Rational and Irrational Numbers: Multiplication, DivisionDate: 10/15/2001 at 07:42:51 From: jess Subject: Rational and irrational numbers I really need help with irrational and rational numbers. The main thing I would like explained is the rules for: irrational multiplied by irrational rational multipiled by rational irrational divided by rational Please please can you help? Thank you so much for your time. Date: 10/15/2001 at 11:31:35 From: Doctor Ian Subject: Re: Rational and irrational numbers Hi Jess, Any rational number can be written as one integer divided by another, right? That's the definition of a rational number. Now, suppose you have two of them, and you multiply them together. What happens? a c a * c - * - = ----- b d b * d Now, if a and c are integers, then a*c is also an integer, right? Similarly for b and d. So the product of two rational numbers must be a rational number. Similar reasoning can show that dividing one rational number by another must also produce a rational result. (Recall that to divide by a fraction, you invert and multiply.) There are no corresponding rules for two irrational numbers. For example, __ __ \| 2 * \| 3 is irrational, but __ __ \| 2 * \| 2 is clearly rational. Also, __ \| 2 ------ __ \| 3 is irrational; but \| 2 ------ __ \| 2 is clearly rational. So far, here's what we know: *,/ rational irrational rational rational ? irrational ? either So what about a rational and an irrational? Well, suppose we multiply a rational number (a/b) by some unknown number (?) and end up with another rational number (p/q). a p - * ? = - b q What can we say about the mystery number? Well, from the definition of division, we know that p - q p b p * b ? = ----- = - * - = ----- a q a q * a - b Now, we know that a, b, p, and q are all integers. So this says that if we multiply a rational number by something, and get a rational number, the something _must_ be a rational number. What this means is that if we multiply a rational by an irrational, we can't end up with a rational product, because to get a rational product, both factors have to be rational. Therefore, the product of a rational and an irrational must be irrational. If you've never seen this kind of reasoning - called 'proof by contradiction' - before, it can seem a little dizzying. But you may as well get used to it now, since you'll be seeing a lot of it in the years ahead. The basic idea is this: You assume that something is true, and then show that this assumption must lead you to conclude something that you know for sure is _not_ true, which means the thing you assumed must be false instead of true. This may be easier to follow if you think about something other than numbers. For example, suppose you're accused of a crime, but there are witnesses who say that you were on the other side of town when it happened. The court would reason this way: Let's assume that you committed the crime. Then you must have been at the scene of the crime when it was committed. But we know that you were somewhere else! Therefore, we know that you didn't commit the crime. Similar reasoning can show that the quotient of a rational and an irrational - or the quotient of an irrational and a rational - must be irrational. So here is our final table: *,/ rational irrational rational rational irrational irrational irrational either Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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