Bit of maths

[Letz]

Originally Posted by Lazlo

Now you've missed the recurring bit as well!

10 multiplied by 0.9 recurring is 9.9 recurring.

Yes but if you're going to use recurring then all the subsequent results are recurring (like Juz says) and therefore none of the results in the OP are correct.

Get a calculator with a 'recurring' button and you'll see

[nigelH]

Originally Posted by Lazlo

Apparently 0.999… (nought-point-nine recurring) is the same as 1. Not just very close, but precisely identical

That's right.
'How do you prove two things are identical?'
'If, for any error, you can prove they are closer than that.'

If you specify a fraction with no matter how many zero between the point and the first significant digit I can get 0.99999.. closer to 1 than that by going one place further. QED.

Worrying about maths is bad for your brain. Russel took half a big thick book to prove that 1+1=2 and yet his proof falls apart because of a flaw in set theory so we can't be sure...

[Lazlo]

Originally Posted by Juz

9.9999999999999r minus 0.99999999999r

=

8.999999999999999r

Juz

Ok, so what I think you're saying, and it's hard to explain, is something like this:

The first number (10 x 0.9 recurring) has one fewer decimal place than the second number (0.9 recurring), so when they are subtracted there is a spare (0.0000000000000....recurring...1) left over.

Something like that.

I think it's one of those paradoxes that appears when dealing with infinity.

[Mark Griffith]

Just give the cashier £10 for your petrol and tell her to keep the change if there is any
cheers
Mark

[David G]

How about this then....

a= 0.99999999999999r

10a = 9.99999999999r but one less r9 than a.......

so when you take a away from 10a you will have an extra 0.000000...........000009 to take away so 9a will equal 8.9999999999999.......99999991

and therefore 9a/9 (or a) will be less than 1 by 0.000000.......000001 or 0.99999r.

You can not multiple 0.999999r by 10 and then say it is 9.99999r as it is actually (r-1) cos you multiplied it by 10

David.

[Lazlo]

Originally Posted by nigelH

That's right.
'How do you prove two things are identical?'
'If, for any error, you can prove they are closer than that.'

If you specify a fraction with no matter how many zero between the point and the first significant digit I can get 0.99999.. closer to 1 than that by going one place further. QED.

Spoken like a true engineer .

Originally Posted by nigelH

Worrying about maths is bad for your brain. Russel took half a big thick book to prove that 1+1=2 and yet his proof falls apart because of a flaw in set theory so we can't be sure...

This is presumably to do with the fundamental axioms of mathematics (of which I seem to remember there are six? Axioms like:

the definition of multiplication is: a * 1 = a
and
the definition of addition is a + 0 = a

I think somebody once proved that even those axioms couldn't be relied upon, and therefore the entire body of mathematical proofs are also therefore suspect. Is this what you'd be referring to?

At school someone told me that parallel lines meet at infinity. Not only that but they meet at right angles. How weird is that!

The other one is to do with asymptotic curves which get closer and closer to a value, but never meet - except at infinity, in theory.

[Lazlo]

Originally Posted by Sixsquid

Yes but if you're going to use recurring then all the subsequent results are recurring (like Juz says) and therefore none of the results in the OP are correct.

Get a calculator with a 'recurring' button and you'll see

Trouble with calculators is they are based on computer chips, which are incapable of expressing even a number like 3 precisely...

[David G]

Originally Posted by Lazlo

Trouble with calculators is they are based on computer chips, which are incapable of expressing even a number like 3 precisely...

Its that Binary thing again isn't it?

There are 10 types of people in the world: Those who understand binary, and those who don't...

David.

[Lazlo]

Originally Posted by David G

How about this then....

a= 0.99999999999999r

10a = 9.99999999999r but one less r9 than a.......

so when you take a away from 10a you will have an extra 0.000000...........000009 to take away so 9a will equal 8.9999999999999.......99999991

and therefore 9a/9 (or a) will be less than 1 by 0.000000.......000001 or 0.99999r.

You can not multiple 0.999999r by 10 and then say it is 9.99999r as it is actually (r-1) cos you multiplied it by 10

David.

Yes, that's along the lines of what I was trying to explain in the previous post to yours. It requires that there are different brands of infinity

[mke]

Recurring 9s go on for ever.
So 0.9 recurring and 9.9 recurring have the same 'number' of 9s after the decimal point.
So when you subtract, the recurring parts (after the decimal point) cancel each other out leaving the exact value of 9

The argument in the OP is precisely correct, and 0.9 recurring is really exactly equal to 1!

There is an algebraic proof based on geometric series ... a little light reading for a deco stop one day? )

Mike