Help with a proof :D

[bver]

Ok, i have to prove that .49... = .5


Here is my basic proof so far.

If .49... < .5

then there has to a number N such that

.49... + N = .5


The problem im having is that i dont know how to prove that for all N

.49... + N > .5

One thought i had was let N = .5 - .49...

But my teacher said, that the subtraction algorithym starts from the right most point, and Since .49... will never have a right most point, he said i coulndt use that.

So anyother help would be gladly appreciated, Thanks!

[Beune]

proving 0.49999...=0.5 is the same as proving that 0.0999...=0.1 or 1=0.999..
you may find several proves of this on the forum at "2=1"
this may be one:
0.999..=[9*infinitesum(1/10^n)]-9 (for n=0) =[9*lim when n->+infinite of (1-(1/10)^(n+1))/(1-1/10)]-9 =1 (it's the formula for a geometrical sum) so it is 9*(1-0)/(9/10)

[Beune]

sorry i forgot to delete "so it is 9*(1-0)/(9/10)" at the end

[jemidiah]

There's an Edit button on your previous post that you can use for things like that

[Didn't PM for education/ease of access]

Welcome to the boards!


Speaking of editing, turns out I have a few ramblings on this subject.

It seems to me that a number such as 0.49999... could only be represented as a limit; that is, as a number itself, you need to think of it as 0.4 + .09999... with the number of 9's approaching infinity. Then, you can apply the infinite sum of a geometric series above to prove that 0.0999... = 0.01.

The reason I say this is that some people discount the above proof because it involves infinite limits; what I'm saying is that 0.49999.... cannot be represented in any way except with the use of limits, so to define 0.49999... must involve and acknowledge that we are really using a limit.


I know my 7th grade Algebra teacher gave a semi-proof using something that ended up being the geometric series, but I can't remember it. Oh well



The proof that you've already started on is logical, but I've never really gotten into any way of working with inequalities algebraically to produce a proof. I wish I could, actually, but I'm not sure how to continue where you've started.

[bver]

This is what my teacher said in regards to limits

The problem with limits, though, is they only tell you what you are getting close to ...
they don't guarantee that you ever actually get there. For example, if f(x) = 1/x, then
lim (x--> infinity) of f(x) = 0, but that doesn't mean that we ever actually get f(x) = 0 ...
in fact, we don't.

( ...unless you want to say that we get there "at infinity", which seems to me a
dangerous sort of thing to say, since we never are "at infinity" either...)

So just because lim (number of decimal places --> infinity) of the difference appears to
be zero doesn't by itself mean that the difference actually IS zero. It does tell us that
the difference is CLOSE to zero (i.e. really small), but not that it actually IS zero.

Now im completely clueless...

[zaza]

Don't worry about limits. Read this thread and feast thereon.

zaza

[bver]

Not that this helps but.


Any repeating decimal can be represented by the ratio of two integers in the following form.

a/b(length of a) , Where A is the number you want to repeat, and b is a number of nines equal to the length of a.

Ex. 33/99 = .333...
1234/9999 = .12341234...
4726214/9999999 = .472614...

After reading that thread i realized that there is only one fration that can represent .9... and that is

9/9 or 1

So .999... = 1

In any event I'm lost on how to prove that you could never add a number N to .49... to equal .5.

My thoughts are that becuase we are talking about an infinite set, there is no number N that you could ever add to .49.. that is small enough that would eqaul .5

Any way thanks for the help

[bilm_ks]

let x = 0.499999....
10x=4.99999......
10x-x=4.5 -> 9x = 4.5 -> x = 4.5/9 = 0.5

One more proof for the illusion of infinite!