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copied!<p>(Was inspired by Tony Breyal's rather good answer to post one of my own)</p> <p>Zero is a tricky and subtle beast - it does not conform to the usual laws of algebra as we know them. </p> <p>Zero divided by any number (except zero itself) is zero. Put more mathematically:</p> <pre><code> 0/n = 0 for all non-zero numbers n. </code></pre> <p>You get into the tricky realms when you try to divide by zero itself. It's not true that a number divided by 0 is always undefined. It depends on the problem. I'm going to give you an example from calculus where the number 0/0 <strong>is defined</strong>. </p> <p>Say we have two functions, f(x) and g(x). If you take their quotient, f(x)/g(x), you get another function. Let's call this h(x). </p> <p>You can also take limits of functions. For example, the limit of a function f(x) as x goes to 2 is the value that the function gets closest to as it takes on x's that approach 2. We would write this limit as:</p> <pre><code> lim{x->2} f(x) </code></pre> <p>This is a pretty intuitive notion. Just draw a graph of your function, and move your pencil along it. As the x values approach 2, see where the function goes.</p> <p>Now for our example. Let:</p> <pre><code> f(x) = 2x - 2 g(x) = x - 1 </code></pre> <p>and consider their quotient:</p> <pre><code> h(x) = f(x)/g(x) </code></pre> <p>What if we want the lim{x->1} h(x)? There are theorems that say that</p> <pre><code> lim{x->1} h(x) = lim{x->1} f(x) / g(x) = (lim{x->1} f(x)) / (lim{x->1} g(x)) = (lim{x->1} 2x-2) / (lim{x->1} x-1) =~ [2*(1) - 2] / [(1) - 1] # informally speaking... = 0 / 0 (!!!) </code></pre> <p>So we now have:</p> <pre><code> lim{x->1} h(x) = 0/0 </code></pre> <p>But I can employ another theorem, called <strong><em>l'Hopital's rule</em></strong>, that tells me that this limit is also equal to 2. So in this case, 0/0 = 2 (didn't I tell you it was a strange beast?)</p> <p>Here's another bit of weirdness with 0. Let's say that 0/0 followed that old algebraic rule that anything divided by itself is 1. Then you can do the following proof:</p> <p>We're given that:</p> <pre><code> 0/0 = 1 </code></pre> <p>Now multiply both sides by any number n.</p> <pre><code> n * (0/0) = n * 1 </code></pre> <p>Simplify both sides:</p> <pre><code> (n*0)/0 = n (0/0) = n </code></pre> <p>Again, use the assumption that 0/0 = 1:</p> <pre><code> 1 = n </code></pre> <p>So we just proved that all other numbers n are equal to 1! So 0/0 can't be equal to 1.</p> <p><em>walks on back to her home over at mathoverflow.com</em></p>

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