[split] 0.999... equals 1

[Tiberius]

Fixed value with a never-ending number. Values never change unless you give them a rate of change. 0.9r has no rate of change, ergo it's value is constant even when the number representing it is endless

[Edwardo Piet]

It's very simple Sae, once you get that we are just talking about an infinite number of 9s on the end of 0.9. We are not talking about the value infinity, we are talking about a string of 9s that go on for infinity on the end of 0.9r. 0.9r= 0.99999999999999999 - the 9s going on forever. And the point is there can be no gap between that and 1, because you would have to put a number on the end of it to make it 1, but there is no number to put on the end of 0.9999999999999r, which is why it is logically a number of equal value to 1, it's 1 represented in decimal.

0.9r is 1 represented in decimal because there can't logically be a gap between it and 1, which means if it's no less or more than 1, then it logically must be the same value as 1. It's 1 in decimal.

Phew, hope that helps.

EvF

[Tiberius]

Just to make a correction to what EvF said, he asserted "0.9r is 1 in decimal" which whilst true, can be seen to imply that 1 isn't already in decimal, which it is.

1 is a representation of the value 1 in decimal.
0.9r is another representation of the value 1 in decimal.

[Edwardo Piet]

Oops, sorry my mistake lol :S

I thought 1.0 is 1 in decimal. I thought without more than 1 digit then it didn't count or something, apologies for my ignorance (extreme ignorance I presume?).

1.0 and 1 are obviously both decimal then. As is 0.9r of course.

EvF

[Tiberius]

With integers, we generally leave off the decimal point.

1 is technically written:

...000001.000000... in decimal, so there are a load of ways of representing it. 1.0, 01.0, 1.000, etc. Most of the time you ignore the extra 0s since they don't really help

[Edwardo Piet]

Hmmm, is it just that they don't help?

- and what's the difference between all those 0s on each end as opposed to just the one 0 on each end?

If the 0s are completely redundant - they aren't needed - then my question is, why is it any more "technically" written with the 0s on each end than without them - than with just as 1?

EvF

[Tiberius]

Every number fits on the number line, which is infinitely long in both directions, with decimal point in the middle.

Each position on the number line can take one of 10 digits (0-9), depending on the value you want to represent. The positions to the left of the decimal point represent the powers of 10 in the positive direction, so every step you take to the left, the value is worth 10 times more than the previous would have been. The positions on the right of the decimal point represent the powers of 10 in the negative direction, so every step you take to the right, the value is worth 10 times less than the previous would have been.

For example:

00001.00000 = 1 (There is a 1 in the first position on the left, so it is worth 1 x 10^0, which is 1. There are no other values in other positions.)

00010.00000 = 10 (There is a 1 in the second position on the left, so it is worth 1 x 10^1, which is 10. There are no other values in other positions.)

00000.10000 = 0.1 (There is a 1 in the first position on the right, so it is worth 1 x 10^-1, which is 0.1. There are no other values in other positions.)

Ok, so let's do a hard one. Combining multiple positions:

00101.00100 = ?

Well, there is a 1 in the third position along to the left, and a 1 in the first position along to the left, and also a 1 in the third position to the right. So:

(1 x 10^2) + (1 x 10^0) + (1 x 10^-3) = 101.001

This is simple base 10 (decimal) arithmetic, but done in it's true form. It's not very good to teach to people just starting out, but once you have to calculate in more bases (like base 2 - otherwise known as binary) it becomes invaluable.

You can see that it works with any number 0 - 9, so:

00823.90000 = (8 x 10^2) + (2 x 10^1) + (3 x 10^0) + (9 x 10^-1) = 823.9

So, the extra zeros are neccessary, since they represent powers of 10 in the decimal system. Another way of writing the above calculation would have been:

(0 x 10^4) + (0 x 10^3) + (8 x 10^2) + (2 x 10^1) + (3 x 10^0) + (9 x 10^-1) + (0 x 10^-2) + (0 x 10^-3) + (0 x 10^-4) + (0 x 10^-5)

However since all of the calculations that have a multiple of 0 in them are equal to 0, there is no point in including them in the calculation.

Some of you are probably wondering about the extra bases, so I'll quickly do an example of binary to decimal conversion. Binary is the base 2 system, which means that for each position on the number line, there can only be 2 values (0-1) rather than 10 (0-9). Not only this, but each step represents a power of 2 (in positive / negative direction).

So the number: 01010.10100 = (1 x 2^3) + (1 x 2^1) + (1 x 2^-1) + (1 x 2^-3) = 10.625 (in decimal).

Easy!

[theblindferrengi]

Holy crap this has become derantis.

[Edwardo Piet]

Well, that was very interesting insofar as I could comprehend it!

Cool that they are required. It looks cool too I mean lol.

EvF

[kılıç_mehmet]

Is it?