# Dividing the random challenge by 0

I decided to make the random post challenge into a series, and the second post is… unfortunately one I made in the blog’s early days, which is both cringe and trash.

By the way, credit to Ben Orlin.

Oh well.

Alright. The question everyone has wondered at least once.

WhY cAn’T yOu DiViDe By 0?????

You see, it’s a fair question. It’s not such a dumb question and it might actually make sense to ask it. So why can’t we?

First off, let’s give an example. Say it’s your birthday, you ordered a hundred pizzas and you’re inviting one of your friends over. How many pizzas would you guys get each? Well… fifty pizzas. That’s a lot, but it’s equal, and it’s fair.

Now let’s say your social circle expanded and you have two friends that you invited over, which is now three people, including you. How many pizzas each? Well, assuming that each pizza is 6 slices, you would each get 33 pizzas and 2 slices, or 33.333… pizzas. That’s less than with only one friend, but it’s still a lot.

Now, let’s skip a few steps. What if we invited 7 friends, namely 8 people? That would be 12.5 pizzas per person.

Skipping all the way to 49 friends, how many pizzas per person? Now, it’s a bit more realistic: 2 pizzas per. We’ve skipped a lot of friends, but it’s fine.

What about 99 friends? Now we’re getting really realistic: 1 pizza per person. If we go any further, we’ll reach decimals. For example, with 199 friends (wow), you’ll have three slices each (assuming the six-slice pizzas), or half a pizza each.

Let’s go downwards a little bit. What if we had just you? What a lonely party. You’d get a hundred pizzas all to yourself. But what if you started a party, then Thanos did the Thanos snap and you disappeared, leaving no one at the party? How many pizzas would each person get? Obviously, with no pizzas, we can’t do much of anything. Therefore, it’s undefined.

In another context, let’s think about multiplication as repeated addition, and division as inverse multiplication. 5 times 3 means 5+5+5, or you may argue that it’s 3+3+3+3+3, but let’s say, to not confuse ourselves, that x times y is x+x+x+…+x+x with y x’s. Let’s take the example of 12. 12 is 3 times 4, or 3+3+3+3. What if we wanted 0 to be one of its factors? If 12 is 0 times x, then 12 = 0+0+0+0+…+0+0 and no matter how large x is, we’ll never reach 12. Therefore, it’s undefined.

The final way, let’s use the same logic, but a little more algebraically. Let’s say that 12/0 was a thing, and let’s call it u. that means, 12/0 = u, and therefore 0 times u is 12, or u times 0 is 12. Now, any number multiplied by 0 is 0, so we can’t reach it. We never will.

Now, an interesting thing.

Let’s say that we could divide by 0. And in limits, it technically is. Say we had the limit as x goes to 0 for 12/x multiplied by x. Obviously, 12 divided by x multiplied by x is like 12 being multiplied by x/x, which is logically 1. Therefore, due to cancellation, it would be 12. But mathematically, the limit is saying (12 * 0) / 0, which is 0/0.

I’ll go into detail on 0/0 in the next post. Until then, peace.

This site uses Akismet to reduce spam. Learn how your comment data is processed.