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My 6-year-old’s homework was “to find the nearest $ 10 $.” For example, $$ 42 \to 40 \quad \text{and} \quad 28 \to 30. $$ For $ 55 $, she answered “$ 50 $” and was marked wrong. How is this wrong? Clearly, $ 55 $ is slap-bang in the middle of $ 50 $ and $ 60 $, so surely either answer is correct. The question does not mention rounding of any sort, therefore you can’t say that common rounding is being used. Comments are most welcome!

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What metric space did the teacher specify? –  Théophile Jun 19 at 15:56
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Thank you for all your answers, very interesting to hear all your opinions. What concerns me is that 6 year olds are being taught that 5 is closer to 10 than it is to zero. Does this mean when they are 5 miles through a ten mile walk they will think that it's further to walk back the way they came than to carry on? I feel like I'm missing something here! –  Lou Jun 19 at 19:40
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@Nit, its unlikely for a 6 year old's homework... –  George Jun 19 at 20:09
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Was it a large amount of "5". So she expected 60... in small amounts of "5" I could understand answering 50... –  jcaruso Jun 19 at 21:58
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The main lesson here is that the teacher likely taught and gave examples of rounding 5 up to 10 in response to questions like this, so the 'correct' answer was to answer in accordance with those examples. This is less a lesson in mathematics and more a lesson in understanding and meeting personal expectations for the purpose of achieving a goal (a good score on the test.) The approach being taught may well be flawed, but being able to tolerate doing things incorrectly for the sake of expediency is perhaps a valuable life skill. Then again, maybe not... –  Dan Bryant Jun 19 at 23:00

5 Answers 5

up vote 17 down vote accepted

$$|55-50|=5 \\ |55-60|=5 \\ 5=5$$ QED

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tongue a little bit in cheek here ;) –  martin Jun 19 at 23:32
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I wish all proofs were so easy. –  TheBluegrassMathematician Jun 20 at 2:49

I would agree with you that both answers should be correct. Despite your plea not to mention common rounding, I suspect that is where the problem is-the answer key was made up with "round 5 to even" or "round 5 up" in mind.

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Also, it seems logical as a convention to have 5 rounded to 10, as if we divide the numbers 0-9 into two halves, it would have to be 0-1-2-3-4 and 5-6-7-8-9 to have the same number of numbers in each half. –  Hippalectryon Jun 19 at 15:43
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Yep, but if the question was formulated as "find the nearest", the question is bogus as there are two nearest. –  Yves Daoust Jun 19 at 15:50
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@Hippalectryon That argument is not correct, because numbers end in 0 are not rounded at all, neither up nor down. So you are rounding numbers ending in digit 1-2-3-4 down, and those ending in 5-6-7-8-9 up, and this does in fact bias the results upward. (To see this, generate ten thousand random numbers, and sum them; then round each one off to the nearest integer, and sum them again. The second sum will invariably exceed the first.) This is why persons who deal seriously with statistics adopt a different rule, such as round-toward-even. –  MJD Jun 19 at 18:32
    
@MJD Ooh that's true –  Hippalectryon Jun 19 at 18:34
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@MJD You should be careful with the word "invariably" in mathematics. There are (very unlikely) cases where the two sums are equal, e.g. when each random number is already an integer. –  Tim S. Jun 19 at 19:33

This is probably an exercise on rounding, where if the one's place is greater than or equal to $5$, then rounding up is considered "correct" (although if you study statistics at a high level, you will find out that this is not the case; there are different rules). However, the question said "find the nearest", in which case both answers are correct. The question in the homework is pretty ambiguous, and IMHO should not be asked.

Getting back to the main question, let's examine the differences between $0$ and $5$, and $5$ and $10$.

$0$ is exactly $5$ away from the number $5$, and so is $10$. The number $5$ is right in the middle of $0$ and $10$. It is not closer to any of the numbers.

$$\underbrace{0, \ 1, \ 2, \ 3, \ 4,}_{\text{5 numbers}} \ \mathbf 5, \ \underbrace{6, \ 7, \ 8, \ 9, \ 10}_{\text{5 numbers}}$$

Imagine a race is held in a $10\text{km}$ straightaway. One car starts at the very left (i.e. at $0$ km) and the other starts at the very right (i.e. at $10$ km). Both cars have to travel $5$ km to the finish line. If they both travel at the exact same speed throughout the race, then obviously they will get there in the exact same time!

Racing on a 10 km straightaway

The point of that was to say that the difference between $0$ and $5$, and $5$ and $10$ are the same! There's no "$0$ is closer to $5$" or any of that stuff.

$5$ is as close to $0$ as it is close to $10$.

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this answer currently violates the answers with pictures get lots of votes principle. curious. –  James S. Cook Jun 20 at 2:14
    
@JamesS.Cook My drawings suck haha –  JChau Jun 20 at 15:16

You're right that it's ambiguous so there is a long-standing convention that you always round the 5 up.

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As the question rightly says, this is not deemed to be a rounding. –  Yves Daoust Jun 19 at 15:52
    
Clearly it is deemed to be a rounding, by the marking scheme. The question is to assemble ammunition to challenge the marking scheme and change how it's deemed ;-) –  Steve Jessop Jun 19 at 23:49
    
It might have just been the case that the question isn't saying what it's really asking. –  Squirtle Jun 19 at 23:49

People like to say that 5 is right in the middle between 0 and 10, but it really isn't. Rather than go into an advanced mathematical explanation, I'll just show it simply. With one digit precision, you have 5 digits that round down: 0 1 2 3 4 and 5 that round up: 5,6,7,8,9. With two digit precision, you have 50 that round down (0-49) and 50 that round up (51-99). If you were to put 5 with 0-4 or 50 with 0-49, you would have an extra number, and would be biased towards rounding down.

And this pattern holds no matter how precise you wish to be. Hence there is actually a reason to choose to always round the exact middle up.

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This is bogus. You don't round 0 to anything. Matter of fact this introduces a bias towards larger numbers because you round 1 2 3 4 down, but 5 6 7 8 9 up. There's a reason anybody who ever had stats 101 will use bankers' rounding (i.e. round half to odd [or was it even? never remember]) if they want a sensible average. –  Voo Jun 19 at 19:21
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I would claim that 5 $is$ right in the middle and that you don't round $0$ at all -- you leave it alone. –  joeA Jun 19 at 19:26
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How do you measure distance, then? 10-5 = 5-0 = 5. If I drive five miles toward something that's 10 miles away, I'm right in the middle between them. I'm not closer to one than to the other. –  Joshua Taylor Jun 19 at 21:43
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... you have five that round down: 0 1 2 3 4, and five that round up 6 7 8 9 10. You say wait a minute 10 doesn't count. I say then neither does 0. –  Joshua Huber Jun 19 at 21:48
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The question isn't about rounding. It's about whether 5 is closer to 0 than to 10. –  David Richerby Jun 19 at 22:51

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