rounding numbers

[richmal]

This is not really homework as I home educate, but it was a question asked me by dd.

If you had .499999(......recurring) at the end of a number, would you round the units up or down?

would you not round up to .5? (is that what you mean)

Down, as 49 is less than 50.

If it were .5099999999999 then round up.
As it is .4999999999 then round down.

If rounding to one decimal place, then round up to .5

So it depends on if it rounding to whole number or not.

Up! Obvs! You'd round .49 up to .5

Definitely round up to 0.5.

I've been doing quite a bit of this today, so I had to revisit the rules. If the last digit is 5, 6, 7, 8, or 9, then round up the second-to-last.

If the last digit is 1, 2, 3, 4, 5, then don't round up.

Continue in that fashion until you reach the required number of decimal places.

Eg:

1.6481745 to 2 decimal places is 1.65

0.26346 to 3 decimal places is 0.264

Thank you for the replies.
I have not written the question clear enough.
Here goes again:
If you had say 1.499999......(I am unable to put a dot over the 9), would it be rounded down to 1 or up to 2, given that 0.49999999999......=0.5
I think round up, DP thinks round down.

Down. You only look 1 place further on, so as it's 1.4.... you round down.

I would say round down to 1 as 1.49999 is less than 1.5.

Down.

If you reach .5 than you round up.

Down to 1 as it's less than 1.5

down.

down, as ilove says you only look at one digit to the right of the one you are rounding to

Down.

But 1.499999recuring=1.5
So if two numbers are the same why is one rounded up and the other down?

you dont do any rounding to start with as you just look at the figures you have and as 1.4999 is less than 1.5 you round down.

I should say the question occurred because we did the proof of 0.999recuring being equal to 1.
So the point is 1.499999......recurring is not less than 1.5, but equal to 1.5.

How is 0.99 recurring equal to 1? It rounds to 1, whatever degree of accuracy you are using, but is infinitesimally smaller than 1, surely?

Ditto for 1.499 recurring - infinitesimally smaller, and therefore rounds down.

The proof goes:
x=0.999.....recurring
10x=9.9999.....
10x-x=9.999...-0.999....
9x=9
x=1

Similarly if x=0.4999...
10x=4.999...
10x-x=4.999...-0.4999...
9x=4.5
x=0.5

The nines go on to infinity so multiplying by 10 means there is still an infinite number of nines.

The answer is still down, as you don't look past the digit 4.

I would say round up.

1.5 which in turns rounds up to 2 if you are not having any decimal places, but that might be an engineering / computing bias.

How is 0.99 recurring equal to 1? It rounds to 1, whatever degree of accuracy you are using, but is infinitesimally smaller than 1, surely?

Because it is nearer to one. If you were booking an event for 99 people and had the choice of 2 venues, one seats 100, the other seats 60 which would you book?

1.4999 is closer to 1 than 2 so I still say round down. You don't round it up to 1.5 first, you look at the actual number and as it is less than 1.5 you round it down.

you have to round down as 1.4999999999 -1 is less than 2-1.499999999.
The rounding up with number ending "5" is just a convention.

Richmal is correct.

Hmm... I get your point now Richmal! It's clearly been far too long since I've done proper maths. Can we go with - any maths involving limits probably doesn't need you to do rounding? (If there is an actual answer I'd really like to know it!)

Actually, having thought abut it a bit I probably agree, Cornflake. It's only for practical purposes you need to do rounding, which is just convention rather than mathematical fact. So I probably am comparing chalk and cheese. (I'm sure there's a mathematician somewhere who has put this a lot better).