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8 solutions
If that's true, why include the ten-thousandths digit at all? It confuses the logic of the question.
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It's probably intended to try and throw people off. The question says to round to two decimal places, so with the 5, somebody (obviously at least 36% that answered this at time of writing) might assume you'd round the 5 up and continue until you reached .50.
Including the ten-thousandths digit is the entire reason the question is even asked. If you know your rounding rules, you will ignore it. If you don't, the inclusion of this digit throws you off, and you get it wrong. The entire point is that you accurately round to the number of significant digits the question asks you to round to. It says two decimal places. So the only number you consider is the digit at the third decimal place. Nothing beyond that.
Out of curiosity, how would we round (to 2 decimal places) 0 . 4 9 4 9 9 9 9 9 9 9 . . . . . , where the 9 's continue to infinity? At first glance the answer would seem to be 0 . 4 9 . but since 0 . 0 0 4 9 9 9 9 . . . . . = 0 . 0 0 5 we have that 0 . 4 9 4 9 9 9 9 . . . . = 0 . 4 9 5 , which we would round up to 0 . 5 0 .
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You're absolutely right Brian
While I agree that 0.0049999... would tend to be very close to 0.005 (in similar vein that 0.9999... is very close to, or equal to 1), I do believe it will be only coming close, but not reaching the exact 0.005, hence I will round these up to 0.49
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The term "0.0049999...." really describes a sequence of numbers - each time you add another nine, you're adding a term to the sequence. So we typically take this value as the limit of this sequence as it approaches infinity. This limit is exactly equal to 0.005.
I think that the key here is that, just as 0 . 9 9 9 . . . is identical to 1 , (and not just "close" to it), 0 . 4 9 4 9 9 9 9 . . . . is identical to 0 . 4 9 5 , so I would round it up to 0 . 5 0 to 2 decimal places.
@Andrew Ellinor What are your thoughts?
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@Brian Charlesworth – Jumping in here @Brian Charlesworth
I agree with what you've said. When we round some x > 0 to n decimal places, we're asking for the number with at most n digits past the decimal point which is closest to x . If two x are equally close, we take the larger one by convention; e.g. we "round up".
With x = 0 . 4 9 4 9 = 0 . 4 9 5 , 0 . 5 0 and 0 . 4 5 are equally close, so we round up to 0 . 5 0 . On the other hand, for any finite number of 9s, we would round x = 0 . 4 9 4 9 9 9 … 9 down to 0 . 4 9 5 .
Of course, the only reason there is even any ambiguity is because 0 . 4 9 4 9 is a terrible way to denote 0 . 4 9 5 .
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@Eli Ross – Haha Yes, I suppose it is the numerical equivalent of an (infinite) run-on sentence. Brevity is highly preferable. :)
@Eli Ross – FTFY: we would round x = 0 . 4 9 4 9 9 9 9 … 9 up to 0 . 4 9 5
Your first glance answer is correct. You are rounding to two decimal places, so the only digit about which you are concerned is the third decimal place. Ignore the infinite 9's. 0.494999999...... Is 0.49, no matter how many 9's there are after that.
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But when there are an infinite number of 9's we have that 0 . 4 9 4 9 ˉ = 0 . 4 9 5 , which we would round up to 0 . 5 0 . So while normally we just have to check the digit in the third decimal place when rounding to two decimal places, we might also need to confirm that the number of 9's after a 4 in the third decimal place is finite. I brought this number up because it appears to be the singular exception to the general rule you mention.
As a person who calculates baseball statistics, I can tell you that the fourth digit after the decimal (a 5) will up the previous number (from 4 to 5) which will up the next two digits from (49 to 50) and this person would have a .500 batting average.
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In that case, you are calculating baseball statistics incorrectly.
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I wrote this without even paying attention. I was paying more attention to the TV. I actually deleted this comment. I read it after I posted it and thought "That's not right", so I deleted it. The person would have a .495 batting average, but this item would be .49 as rounding up from the number 4 is not something you can do. I am just frustrated that somebody read my comment. I felt like such a fool when I read what I had posted.
Just use the method of significant digits of physics to round off any number
Why not? you round up, you round up. Yes 0.4945 is closer to 0.49 than 0.5, but following the rules of maths not math- Iwas taught, 5 and above you round up- maybe things have changed?
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This is exactly what I was thinking. The logic works out to 0.49, but the math works out to 0.50 using standard rounding principles.
But one solely considers the digit in the thousandths place in this case, since we are asked to round to the nearest hundredth. Generally, we pay attention to only the ( n − 1 ) th digit when asked to round to the nearest n th digit.
Hmm... Makes sense. I don't get what's being tested here though!
When rounding, we round to the closest value. So we round up if that's a smaller difference, and we round down if that's a smaller difference.
What happens if the difference is the same in both directions? Here, mathematicians have decided that we should round upwards. There are other choices for what we could do, but rounding upwards is simply the standard way to do it.
So how do we round 0.4945 to two decimal places? We round it down, because subtracting 0.0045 is a smaller jump than adding 0.0055. So the answer is 0.49.
If we split this up into multiple steps, we get the wrong answer. You do not round up to 0.495, and then round up again to 0.50. This number is further from 0.4945 than 0.49 is, so it's a less accurate rounding.
This demonstrates that it's dangerous to blindly follow the steps without understanding what you're doing. Think about what rounding actually "is", and what it's intended to do.
"Here, mathematicians have decided that we should round upwards"
Not if you're dealing with data sets. Or anything in science really. With respect, I'm not sure where you're getting the idea that "mathematicians" have decided on the convention that we should round up from half values. It depends on the work you're doing.
If you adopt this convention with a set of data, you will skew your data upward. The convention in science is to round to the even number if you have to drop a decimal place:
0.45 --> 0.4
0.55 --> 0.6
The reason for this convention is that if you're dealing with large sets of numbers or data points, the rounding errors will cancel each other out.
The number rounding process is as follows:
Let X.X...WY a real number to be rounded to X.X....Z (one less digit) IF Y>=1 and Y<=4 then Round to X.X....Z where Z = W IF Y>=6 and Y<=9then Round to X.X....Z where Z = W+1. (Follow the process for correct representation of the number). IF Y=5 then {IF W is odd then Round to X.X....Z where Z = W+1 (Follow the process for correct representation of the number).} {IF W is even then Round to X.X....Z where Z = W} ENDIF
When rounding from 5, always round even. Therefore, 0.4945 is rounded to 0.494, and further rounded to 0.49. If the thousandths place were an odd number, say 0.4955, then we would round it even to 0.496, and further round to 0.50.
Then what would 0.4946 give when rounded to two digits behind the decimal point?
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It would be the exact same answer, 0.49. The ONLY digit you consider when rounding, is the digit preceding the one you are rounding to. Whether that "6" were a 1, 2, 5, 7, 9, would not matter. If you are rounding to two decimal places, the next number, the third decimal place, is ALL you consider.
I agree with Mark Jackson. We were always taught to round to even. In some cases this would be to round up ; other times it would be to round down.
0.4945 = 0.494 because 3rd place should be odd to get incriment otherwise will be same. So now 0.494 = 0.49
There are rules attached with the rounding off of 5. They say that if and even number is before 5, the 5 changes to 0. And if an odd number is before 5, it rounds off to 10. So, 0.4945 rounds off to 0.494 and then to 0.49 But, 0.4935 rounds off first to 0.494 and then to 0.49 There is a difference between the two statements.
It ALL depends on which decimal place to which you are asked to round the number. The question says two (2) decimal places. So the ONLY digit you need to consider the the third decimal place, in order to determine your rounding. Anything to the right of the decimal place to which you are asked to round, is irrelevant. It's as simple as that.
Acc. to rounding off rule if 5 is on the last then it gave a Successor for every 2nd last digit is odd and vice-versa.......
You will have to take account of the third position. As 4 < 5, the number has to round down to 0.490, or simply 0.49. One cannot just round the fourth 5 to 0.495 then round it again to 0.50.