What is the largest number less than 1?

[Nebuchadnezza]

Sorry for not reading through all of the pages...

But can't you simply prove that $$0.999...$$ is equal to $$1$$ in this matter?

$$0.999...$$ can be expressed as a geometric series. As such

$$\frac{9}{{10}} + \frac{9}{{{{10}^2}}} + \frac{9}{{{{10}^3}}} + ... + \frac{9}{{{{10}^n}}}$$

An infinite geometric series is an infinite series whose successive terms have a common ratio.

The formula for the sum of a infinite geometric series is as follows: $$\frac{{{a_1}}}{{1 - k}}$$ where $$k$$, is the ratio between the terms and $$a_1$$ is the first number in the sequence. Plugging in everything we have. We now get.

$$= 0.999...$$

$$= \frac{9}{{10}} + \frac{9}{{{{10}^2}}} + \frac{9}{{{{10}^3}}}+...+\frac{9}{{{{10}^n}}}$$

$$k = \frac{{{a_n}}}{{{a_{n - 1}}}} = \frac{{\frac{9}{{{{10}^n}}}}}{{\frac{9}{{{{10}^{n - 1}}}}}} = \frac{9}{{{{10}^n}}}:\frac{9}{{{{10}^{n - 1}}}} = \frac{9}{{{{10}^n}}} \cdot \frac{{{{10}^{n - 1}}}}{9} = \frac{{{{10}^{n - 1}}}}{{{{10}^n}}} = {10^{n - 1}} \cdot {10^{ - n}} = {10^{\left( {n - 1} \right) - n}} = {10^{ - 1}} = \frac{1}{{10}}$$

$${a_1} = \frac{9}{{10}}$$

$$S = \frac{{{a_1}}}{{1 - k}} = \frac{{\frac{9}{{10}}}}{{1 - \frac{1}{{10}}}} = \frac{9}{{10}}:\frac{9}{{10}} = \frac{9}{{10}} \cdot \frac{{10}}{9} = 1$$

 

[camilus]

What about in the set of rational numbers Q which is countable

$$\mathbb{Q}$$ is a densely ordered set, which means that for any x and y such that x < y, the exists a z in $$\mathbb{Q}$$ where x < z < y. So there in no greatest number less than 1 because whatever number x you give me when y=1, I can always find a z.

 

[olivermsun]

$$\mathbb{Q}$$ is a densely ordered set, which means that for any x and y such that x < y, the exists a z in $$\mathbb{Q}$$ where x < z < y. So there in no greatest number less than 1 because whatever number x you give me when y=1, I can always find a z.

Just to connect this to the discussion above, also notice that every number in the series 0.9, 0.99, 0.999, 0.9999, ... is a rational number, so there are your arbitrarily close rationals right there.

 

[Thinker8921]

Could someone please clear a doubt I have with infinity being taken as a value.
I would like to start from the beginning to be clear. The sum of a geometric series is given by:
S=(rB-A)/r-1
Where S=sum of the series, r= ratio between terms, B=last term, A=first term.
Now B=A(r^N-1). So,
S=[{rA(r^N-1)}-A]/r-1
i.e, S=[{(r^N)-1}A]/r-1
Now in the case of the infinite geometric series 0.99… ,
N= infinity A= 0.9, r= 0.1
Then there is as the 1st term 0.9, 2nd term is 0.09, 3rd is 0.009 etc. Then the sum of this series will be 0.99...
Also in the numerator is 1-(r^N). This is instead of (r^N)-1. And in the denominator it is 1-r instead of r-1. This is because r is less than 1.
So finally there is:
S=[{1-(r^q)}A]/1-r
Where q represents infinity.
Now since there is infinity as the power of r, this is seen to become:
S=A/1-r
But the problem is it correct to take infinity as a value and approximate like this in the case of 0.99… ?
I thought infinity never had a definite value, that it was instead just an ever-rising concept. Could this 'problem' make the 0.99… equal 1?
How can 0=1/q, so q*0=1 be explained then? (q= infinity)
*I don't know how to write the equations with symbols yet. So sorry for the messy work.

 

[olivermsun]

The infinite sum (if it converges) is defined as the limit of the partial sums
$$\sum_{n=1}^\infty{a_n} = \lim_{N\rightarrow\infty} \sum_{n=1}^N{a_n}.$$

Thus in your expression you take the limit where the exponent goes to infinity and call that the value of the infinite series. This is different from stating that the exponent is actually infinity.

 

[Thinker8921]

Thanks for clearing that up. I got it now.

 

[dalcde]

If you use hyperreal numbers, it might make sense that the largest number is (1-epsilon), where epsilon is an infinitesimal. However, according to my understanding, there isn't a one unified infinitesimal. An infinitesimal can be smaller than another in some weird sense.

 

[Hurkyl]

If you use hyperreal numbers, it might make sense that the largest number is (1-epsilon), where epsilon is an infinitesimal. However, according to my understanding, there isn't a one unified infinitesimal. An infinitesimal can be smaller than another in some weird sense.

Hyperreals satisfy the same theorems (interpreted internally) that the reals do -- there is no largest hyperreal smaller than 1.

e.g. (1 - epsilon) < (1 - epsilon / 2) < 1