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I have a measurement $a=8.3$cm to 1 d.p. I'd like to find the upper bound of this. Why is the upper bound $8.35$cm? not $8.3\dot{4}$cm? Surely 8.35 rounds to 8.4cm? And the lower bound $8.25$cm not $8.24\dot{9}$cm?

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  • $\begingroup$ Then what about $8.3499999999$? $\endgroup$
    – njguliyev
    Commented Sep 1, 2013 at 15:30

3 Answers 3

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The upper bound is 8.35 cm because anything less than 8.35 will round to 8.3, but anything 8.35 or greater will round to 8.4. 8.35 itself does not round to 8.3, but it is the smallest such number, or the least upper bound.

Similarly, the lower bound is 8.25 because anything less than 8.25 rounds to 8.2, and 8.25 rounds to 8.3, so 8.25 is the lower bound (And also the minimum).

Also, your question about $8.25$ and $8.24\dot9$ doesn't really make much sense -- those two decimals represent the same real number.

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  • $\begingroup$ ie $a\in(8.25,8.35)$? But am I not looking for $\displaystyle a\in[a_-,a_+]$ $\endgroup$
    – HCAI
    Commented Sep 1, 2013 at 16:00
  • $\begingroup$ I'm not sure what you mean by $[a_\_,a_+]$. The numbers that round to 8.3 will be $a \in [8.25, 8.35)$, if that helps. $\endgroup$
    – qaphla
    Commented Sep 1, 2013 at 16:48
  • $\begingroup$ Yes that's clear me, but my 15year olds don't know what that notation means. They're asking in essence: 8.35 does not belong to the set of values that round to 8.3. So why should I write it as being an upper bound? The question perhaps boils down to my understanding of the words 'upper bound'. the highest value which will round to 8.3, i.e $[8.25, 8.3\dot{4}]$? Hence it comes with the caveat that I must write $a\in[8.25,8.35)$? $\endgroup$
    – HCAI
    Commented Sep 1, 2013 at 17:11
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    $\begingroup$ The upper bound is simply the value $x$ such that any value that rounds to 8.3 is less than $x$. The issue here is that there is no maximum value which rounds to 8.3. If you give me some value $y < 8.35$ that rounds to 8.3, and claim that it is the maximum, I can always disprove it by taking the value $\frac{y + 8.35}{2}$, which is greater than y, but still less than 8.35, and so still rounds to 8.3. While there is no maximum, there is still an upper bound, because 8.35 is greater than all of these values. $\endgroup$
    – qaphla
    Commented Sep 1, 2013 at 17:17
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The measurement could have been $3.3487$. That would have been rounded to $3.3$, so we cannot conclude that the original measurement was $3.34$ or below.

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  • $\begingroup$ I don't understand this. Presumably 8.3487? Surely that rounds to 8.35? And subsequently to 8.4? ie $a\in(8.25,8.35)$ But am I not looking for $a\in[a_-,a_+]$ $\endgroup$
    – HCAI
    Commented Sep 1, 2013 at 15:59
  • $\begingroup$ Sorry, bad typo. I intended $3.3487$. $\endgroup$ Commented Sep 1, 2013 at 16:02
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Its because the upper bounds of 8.3 is 8.349999999999.... So you round it up to 2 decimal places which is 8.35.

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