I keep stumbling over bounds and error intervals when a value is rounded, especially the inclusive vs exclusive endpoint. For example, if a length is written as 8.4 cm rounded to 1 decimal place, what exactly should the interval be? Is it 8.35 ≤ L < 8.45, or do I include 8.45 as well? And does writing 8.40 to 2 dp change anything about the endpoints? I have lost marks before for using ≤ at both ends, so I am trying to understand the rule in a way that sticks. I feel like I am missing one key idea about which values would actually round to the stated number. Related to that, for a simple sum: if two sides are each given as 3.2 m to 1 dp, what is the tightest possible error interval for their total length? Do I just add the bounds for each, or is there a more careful way to do it? I am probably overthinking this, but I would appreciate a clear way to decide the endpoints in these cases.
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3 Responses
I can’t bring myself to give you a wrong rule here-the real trick that finally stuck for me is: for rounding to the nearest unit of size h (like 0.1 or 0.01), the error interval is half-open: include the lower endpoint, exclude the upper. So “8.4 cm to 1 dp” means 8.35 ≤ L < 8.45, because 8.35 rounds to 8.4, while 8.45 rounds up to 8.5 and must be left out. If you write 8.40 to 2 dp, you’ve changed the precision to hundredths, so the interval tightens to 8.395 ≤ L < 8.405 (same inclusive-left, exclusive-right logic). A memory hook I use-after getting a red pen scolding for writing ≤ on both ends in a test-is “the fencepost belongs to the house on the left”: the boundary point gets included by the value it rounds to, not the one above it. For the sum: if each side is 3.2 m to 1 dp, each lies in 3.15 ≤ s < 3.25, so the total is 3.15 + 3.15 ≤ T < 3.25 + 3.25, i.e., 6.3 ≤ T < 6.5. That’s the tightest possible interval; you don’t need anything fancier than adding the corresponding bounds and keeping track of which ends are open or closed. A nice walkthrough on bounds and rounding lives here: https://www.bbc.co.uk/bitesize/guides/zgsg87h/revision/2.
Think “all numbers that would round to the one shown”: with ordinary half‑up rounding, 8.4 to 1 dp means 8.35 ≤ L < 8.45 (since 8.45 would round to 8.5), and 8.40 to 2 dp means 8.395 ≤ L < 8.405. For the sum, just add the intervals-each 3.2 is [3.15, 3.25), so the total is 6.30 ≤ T < 6.50-pretty sure that’s the tightest unless your course uses tie‑to‑even.
Rule of thumb: when rounding to a last-place size u, the true value is in [stated − u/2, stated + u/2) – include the left end, exclude the right, because that right-end value would round up.
So 8.4 (1 d.p.) means 8.35 ≤ L < 8.45, 8.40 (2 d.p.) means 8.395 ≤ L < 8.405, and two sides of 3.2 (1 d.p.) give a total 6.30 ≤ T < 6.50; that’s the tightest interval, barring some quirky convention.