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As I was looking through Spotify, I noticed that I listened to 2040 minutes of music this year. I did as follows: 2040=204*10, and 60=6*10. Thus 2040/60= 204/6. Intuitively, I'm not able to see why both of these actually equal each other. I mean, the closest I'm able to get to an answer is to assume that this division problem is a ratio, and that I'm trying to find how many minutes listened to 1 "real" minute. However, this makes no physical sense. Can someone please provide some intuition?

Thanks

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closed as unclear what you're asking by Lord Shark the Unknown, Morgan Rodgers, Brian Borchers, user10354138, Cesareo Dec 9 '18 at 9:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Are you trying to count the number of hours you spent listening to music? Very often, the intuition comes from the source of the question $\endgroup$ – Adam Cartisano Dec 9 '18 at 3:38
  • $\begingroup$ Yep, but the main question I'm asking is why we can cancel factors in the numerator and denominator in a division problem. The answer makes sense when we think about fractions as ratios, but not when I think of how many sets of 60 can go into an set of 2040. $\endgroup$ – Dude156 Dec 9 '18 at 3:40
  • $\begingroup$ My mind is totally blown that you are doing contest math with limits, etc. and yet this is somehow lacking intuition. Is this a troll post? $\endgroup$ – Morgan Rodgers Dec 9 '18 at 3:54
  • $\begingroup$ @MorganRodgers No nothing of the sort. For the past two-three months, I've started thinking about math very deeply. It started with asking why we write dx, even though dy is also approaching 0 and we get a different answer (when doing integrals). It then progressed to the distributive property which made sense through area. However, I can't this question. $\endgroup$ – Dude156 Dec 9 '18 at 4:11
  • $\begingroup$ To the people trying to close this question, plz explain why. I'm not asking some sort of troll question. I'm legitimately wondering this. If you recommend that I see a doctor because I might have a had a stroke or something, then please say that instead. $\endgroup$ – Dude156 Dec 9 '18 at 4:24
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Here is a little background before a rather involved answer.

The best question to ask in approaching this problem is "What is division, really?" Not very intuitive, but the short answer is that its the opposite (or, inverse operation) of multiplication. I ask, "What is $2040$ diviced by $60$ equal to?" And in mathematical notation, the answer to that question can be given a name: $x$. This translates to the familiar

$\dfrac{2040}{60} = x = 34$.

The rules of math give us an equivalent question to ask, "What number do I multiply by $60$ to obtain $2040$?" The answer to this one is exactly the number of piles of $60$ objects required to have $2040$ total objects. That number is $x$!

If you like to organize your piles into rows, say, rows of length $60$ all stacked like pancakes, then $x$ is the number of rows, or the height of the piles. Your question is why (in a physical kind of way) can I first cancel factors from the numerator and denominator?

Here is my answer. I intentionally did not state "piles of what objects," because what if I had $204$ dimes, and I broke them into piles of $6$ dimes each? I would have $x$ piles of dimes when I'm finished. But what if I replace the dimes with sets of $10$ pennies? there would still be $x$ piles, I started with the same amount of money, but each pile now has $60$ coins in it!

The more abstract answer to your question is that the physical equivalent to canceling factors in both the numerator and denominator of a ratio is like viewing the numerator as a smaller collection of bigger objects, and breaking it into smaller piles of these bigger objects. In this case, "Bigger" means "each object contains the common factor number of the original objects."

In your example, viewing $\dfrac{2040}{60}$ as $\dfrac{204}{6}$ is like taking $2040$ minutes and breaking them into $34$ piles of $60$ minutes each, or taking $204$ ten minute blocks, and breaking all $204$ of them into $34$ piles of $6$ ten minute blocks. This is the same as taking $34$ sixty minute blocks and breaking them into piles of one hour, which gives one block per pile as desired.

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  • $\begingroup$ Great answer! How would you go about making it rigorous though? $\endgroup$ – Dude156 Dec 9 '18 at 4:36
  • $\begingroup$ Thanks! I would probably do something like, let $x=\dfrac{a}{b}$ such that $a,b\in \mathbb{Z}$ for example, and let $p\in\mathbb{Z}$ divide both $a$ and $b$. Then $\dfrac{a}{b}$ can be written as $\dfrac{pc}{pd} = \dfrac{p}{p}*\dfrac{c}{d}$ for integers $c,d$. Then turning the wheel I would have $c$ lots of $p$ units divided by $d$ lots of $p$ units, and after cancelling the problem returns to its original state. $\endgroup$ – Adam Cartisano Dec 9 '18 at 4:50
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How many sets of 6 go into 204? Ten times as many go into 2040. You need 10 groups of 6 to have one group of 60. We calculated the group's of 6 from the previous step so just divide by 10 to get the group's of 60.

To get how many times 60 goes into 2040 requires finding how many times 6 goes into 204, times 10, then divided by 10.

More generally:

Let us have ratio a/b=n, or how many sets of b go into sets of a. Then c times n is how many sets of b go into sets of ca. We know how many times a set of size b goes into ac, nc. How many times does a set of size c times bigger than b go into nc? It has to go in c times fewer. So we divide nc by c to get n.

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  • $\begingroup$ Solid way of looking at it. Thanks man! $\endgroup$ – Dude156 Dec 9 '18 at 4:32
  • $\begingroup$ How would you go about making it rigorous though? $\endgroup$ – Dude156 Dec 9 '18 at 4:36

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