0 As the DenominatorDate: 01/19/2002 at 17:20:43 From: Kyle Stout Subject: 0 as the denominator Can zero over zero equal anything? My brother, who is a freshman in college, says it can, but my math teacher says it can't. At first I agreed with my teacher, but my brother showed me this whacked-out equation that made it look as if zero over zero can equal anything. Even so, I told him that anything divided by zero is undefined and can't be done, but he stayed with his idea. My math teacher recommended I write you this question. Thank you. Date: 01/19/2002 at 17:58:24 From: Doctor Ian Subject: Re: 0 as the denominator Hi Kyle, A lot depends on what you mean by 'zero over zero'. If you're trying to divide something by the constant value zero, you're right - it can't be done. The operation is undefined. The 'whacked-out equation' that your brother was showing you probably had to do with limits, that is, with predictions about what the ratio of two functions would be if you looked at their values at infinity - which, of course, you can't. Let's say you have a function like f(x) = 1/x. As you get farther from the origin, the value of the function gets closer to zero: f(1) = 1 f(2) = 0.5 f(10) = 0.1 f(100) = 0.01 f(1,000,000,000) = 0.000000001 In fact, although the function never actually reaches zero, we can make it get as close as we want by going far enough out. So we say that 'the function goes to zero as x goes to infinity'. But that's a lot to write, so we abbreviate it limit f(x) = 0 "The limit of f(x) as x approaches x->inf infinity is zero." Now, suppose we have a second function, g(x) = 2/x. This also goes to zero as x goes to infinity. So we have limit g(x) = 0 x->inf Here's the fun part. What if we divide one limit by the other? On the one hand, this looks like limit f(x) x->inf 0 ------------ = - Undefined? limit g(x) 0 x->inf But that's somewhat misleading, because of course, neither the numerator nor the denominator ever really _gets_ to zero. In fact, for any value of x that we choose, it's pretty clear that g(x) will be twice f(x), so the ratio is really 1/2. If you think about this for a while, it should become clear that you can, indeed, come up with two functions, the ratios of whose limits would be any number you want. (Would you like the ratio to be 12.3? Okay, let's let f(x) = 12.3/x, and g(x) = 1/x.) So I would say this: division by _zero_ is undefined, but division by a _limit_ that goes to zero is indeterminate (that is, the value you get depends on the limits you start with). As long as you're careful to make the distinction, there shouldn't be any confusion. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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