Err no. 0 would be a shift from 1/2 to either left or right depending on what direction you wanted to use. 1 maps to a left or right shift that would be the opposite of 0. This means that -1 would map to either -1/4 or -3/4 depending on what you did

I don't know the derivation for that one, just looked on the wiki page for series expansion of root 2. How about:

$e^x = \sum_{i=0}^{\infty}\frac{x^i}{i!}$

sub in $x=1$.

You can do that one yourself. Derive the Taylor series of $e^x$, show the radius of convergence includes $x=1$ (it's actually infinite, the function is analytic). Sub in $x=1$. All the finite sums are rational. The infinite sum is the irrational $e$.

No you have a valid point, however if I'm just halving distance between 2 rational numbers I'm not summing them like those sequences all do. Any number 1/2 way between 2 rational numbers will always be a rational number. I can just sum those 2 numbers and then divide by 2 and that will give me a new rational number

Err no. 0 would be a shift from 1/2 to either left or right depending on what direction you wanted to use. 1 maps to a left or right shift that would be the opposite of 0. This means that -1 would map to either -1/4 or -3/4 depending on what you did — Umonsarmon

Maybe I've got it wrong but here's what my understanding is

1. You convert all reals into binary so that we have only 1's and 0's
2. You need direction to distinguish between 1's and 0's. Left or right depending on what the digit is, 0 or 1.

Now I want you to follow your scheme and map the following numbers

a) 1

b) -1

c) 0

These numbers are simple because they only require 1 step i. e will map to 1/2. There are two types of 1/2. One is +1/2 and the other is -1/2 and we have 3 numbers to consider +1, -1 and 0. It's impossible.

With this procedure you can produce rationals arbitrarily close to irrationals but not irrationals. If you want the procedure to be infinite, you're no longer dealing with rationals (since the series doesn't converge in the rationals, it only converges in the reals).

Ok here we go remember we start at 1/2 and then shift either left or right by half. Lets say that 1 is left and 0 is right.

1 goes to 1/4 (i.e halfway between 0 and 1/2
-1 just gets represented as the key -1/4
0 right shifts from 1/2 to 3/4

To be fair i will have to give that some thought as to whether I can produce a proof that they all terminate on rational numbers. Not an easy task I know. However the total length travelled by an irrational across the line is equal to a 1 if its an irrational number because each time we move between two points we are shifting the point by some 1/2^y. All those sums are rearrangments of the sum 1/2 + or - 1/4 + or -1/8 and so on. Now if the sum of the distance is always 1 which we get by simply measuring the length travelled regardless of whether we are left or right shifting then then the sum breaks down to 1 = (sum of positive fractions - sum of negative fractions). Both these values will be rational numbers because all they consist of are fractions of the form 1/2^y. .

Ok here we go remember we start at 1/2 and then shift either left or right by half. Lets say that 1 is left and 0 is right.

1 goes to 1/4 (i.e halfway between 0 and 1/2
-1 just gets represented as the key -1/4
0 right shifts from 1/2 to 3/4 — Umonsarmon

It would've been easier to say 0 maps to 0 but that would break your rule or demands tweaking it a little bit.

If you graph your rule the output of the function (1/2)^x approaches zero which raises the problem that two real numbers 0 and infinity map to zero. Do you find that problematic?

Then your list is not a list of all the real numbers. There is no question that there are as many rationals as there are naturals. It’s only when you add in all the irrationals to make the reals that you get a larger infinity.

Let me just clarify my proof to fdrake. The sum breaks down to 1 = 1/2 + sum of positive fractions - sum of negative fractions. The sum of the positive fractions is equal to a rational number and hence the negative fractions are also rational. Why is this the case, well here goes.

If we have the sequence 1/4 + 1/8 + 1/16 etc etc no matter how many numbers I remove from that list the product will be rational. If we sum them all up then the product is 1/2. If the sum of the total is rational than any total in the sequence at any point also has to be rational. If I knock out numbers from that sequence than that still leaves me with a rational total otherwise I could add up those numbers in a different order and somehow produce an irrational number. This is impossible from that perspective. If the sum of the total is rational than how can we subtract an amount from this to create an irrational number if the only numbers we are subtracting are rational numbers. So no I stand by my proof ;)

If we have the sequence 1/4 + 1/8 + 1/16 etc etc no matter how many numbers I remove from that list the product will be rational. — Umonsarmon

$e^x -1 = \sum_{i=1}^{\infty}\frac{x^i}{i!}$

No.

Well here goes again

The sum is 1/2 + sum of positive fractions - negative fractions

The maximum sum of the positive fractions is a 1/2 from the sum 1/4 + 1/8 +1/16 etc

Now if the sum is a rational number then at no point in the sequence 1/4+1/8 etc can the sum be an irrational number otherwise the sum of the whole sequence will be irrational.

Now I can rearrange this sequence in all possible combinations and at all points in all these rearrangements the sum is rational. If it were not then the product would be irrational ok.

Now I can arrange the sequence 1/4+1/8+1/16 in such a way that the positive fractions in our sum are in a sequence. But this sequence has to sum to to a rational number from what was argued earlier. This means that the sum of the positive fractions is always rational...Hence the negative fractions are also rational because the produce is a 1/2.

Now if the sum is a rational number then at no point in the sequence 1/4+1/8 etc can the sum be an irrational number otherwise the sum of the whole sequence will be irrational. — Umonsarmon

No.

1 + root(2) + 1/2 - root(2) + 1/4 + 1/8 ...

Also, if your sequence must output a rational, it can't output an irrational.

del

So at what point is a number of the form 1/2^y an irrational number.? I'm curious. The proof works if you understand it, what you've suggested is complete nonsence with regards to the sum you posted

I am confused. Cantor agrees and shows that the rationals are countable. As far as I can understand you, he uses a similar method to your own, except he does not have recourse to binary.

But what is uncountable is the reals, which include irrational numbers like Sq root 2, which was discovered to be irrational by the Ancient Greeks, much to their consternation. So where are root 2 and pi in your list?

The proof works if you understand it, what you've suggested is complete nonsence with regards to the sum you posted — Umonsarmon

(1) The sequence contains irrationals. The infinite sum remains rational.
(2) The sequence can consist only of rationals. The infinite sum can be irrational.
(3) The sequence can consist only of rationals, it can be strictly increasing or decreasing, but not converge in the rationals. (see 2)
(4) The sequence can consist only of irrationals, it can be strictly increasing or decreasing, but converge to a rational.

You don't have the sequence/series facts down.

0.1 would be 1/2 to the right and 1/2 to the left to give us 1/4

1.1 would be 1/2 to the right and again 1/2 to the right to give us, again, 1/4

Both 0.1 and 1.1 are mapping on to the same fraction 1/4.

There is no bijection and the proof is faulty.

1.1 would give the number 11 in binary. This means we either shift twice to the left or twice to the right depending on what direction you give the 1,s and 0,s . Lets suppose its left. We start at 1/2 and left shift to 1/4 then we left shift again to 1/8

-1.1 would just be represented by the key -1/8

0.1 would shift right for zero going to 3/4 and then shift left to halfway between 3/4 and a 1/2

i.e A=0 B=1/2 C=1

This goes to point D halfway between B and C

A=0 B=1/2 D=3/4 C=1

The next shift is left so we have
A=0 B=1/2 E=5/8 D=3/4 C=1

So it terminates on 5/8

0.1 would be 1/2 to the right and 1/2 to the left to give us 1/4

1.1 would be 1/2 to the right and again 1/2 to the right to give us, again, 1/4

Both 0.1 and 1.1 are mapping on to the same fraction 1/4.

Bijection fails.

(1) The sequence contains irrationals. The infinite sum remains rational.
(2) The sequence can consist only of rationals. The infinite sum can be irrational.
(3) The sequence can consist only of rationals, it can be strictly increasing or decreasing, but not converge in the rationals. (see 2)
(4) The sequence can consist only of irrationals, it can be strictly increasing or decreasing, but converge to a rational. — fdrake

Lets go through this proof again.

We have the sum 1=1/2 + sum of positive fractions - sum of negative fractions

So where does the 1 come from you might ask. Well if you take all shifts as positive regardless of direction then the distance the line travels = 1/2 +1/4 +1/8 etc which sums to 1.


Now this is equal to 1/2 + sum of the positive fractions - sum of negative fractions ok.

This gives 1/2 = sum of positive fractions - sum of negative fractions.

Now the sum 1/4 + 1/8 + 1/16 etc sums to a 1/2

I hope you would agree that no matter how I reorder this sum then it always = 1/2 ok
Also no fraction of the value 1/2^y will be irrational on its own
All numbers in the sum are rational
Now at all points in that sum the sum is always rational at any point we sum up to regardless of the order of the numbers. If this were not the case then the sum would not be rational.
If at all points in all sequences of this sum the product is rational then we can find a particular ordering with only the positive fractions being summed which will be rational from what has just been argued and with the remainder of the sequence being the negative fractions which must also sum to a rational value in order for the product (1/2) to remain the product.

Hence all irrational numbers are mapped to rational keys and my proof stands.

0.1 would be 1/2 to the right and 1/2 to the left to give us 1/4

1.1 would be 1/2 to the right and again 1/2 to the right to give us, again, 1/4

Both 0.1 and 1.1 are mapping on to the same fraction 1/4. — TheMadFool

01 goes first right which means it goes 1/2 the distance between 1/2 and 1 which are points on a line to first give 3/4. Then it left shifts halfway to the next point which is 1/2. This means it sits halfway between
1/2 and 3/4 which is 5/8. This value is halfway between 1/2 and 3/4. This is why I would normally use a square with a line going halfway down the middle. Each shift creates a new line and you only move halfway to your next line ok

There isn't a third possibility here. — Wittgenstein

Yes there clearly is a third possibility, it's not a case of either Cantor is right, or the op is right, because both Cantor and the op might be wrong.

… like Sq root 2, which was discovered to be irrational by the Ancient Greeks, much to their consternation. — unenlightened

Actually I think it was Pythagoras who first proved that the square root of two is irrational. And I think it really frustrated him because it demonstrates that the geometrical figure, the square, cannot be a real figure, The two perpendicular sides of a square are incommensurable. This is similar to the problem with pi. It appears like there is an incommensurability between one dimension and another which makes two dimensional figures inherently irrational. It's very odd when you think about it because it casts doubt onto our understanding of spatial existence in terms of dimensions. It may well be that spatial existence would be better represented in another way.

Now this is equal to 1/2 + sum of the positive fractions - sum of negative fractions ok.

This gives 1/2 = sum of positive fractions - sum of negative fractions.
Now the sum 1/4 + 1/8 + 1/16 etc sums to a 1/2
I hope you would agree that no matter how I reorder this sum then it always = 1/2 ok — Umonsarmon

Just looking at this it appears you are rearranging a conditionally convergent series and expecting the same sum. If this is the case you can draw no logical conclusion.

https://en.wikipedia.org/wiki/Riemann_series_theorem

But I haven't followed the discussion so I may be misinterpreting your argument. If so, I apologize.

Please tell me how you post mathematical expressions on this forum. Thanks.

Here..!

$\int\limits_{0}^{1}{\phi (z,t)dt={{\lambda }^{2}}}$

Aha! The Wikipedia cut/paste on Mathtype with square vs angle "math"

01 goes first right which means it goes 1/2 the distance between 1/2 and 1 which are points on a line to first give 3/4. Then it left shifts halfway to the next point which is 1/2. This means it sits halfway between
1/2 and 3/4 which is 5/8. This value is halfway between 1/2 and 3/4. This is why I would normally use a square with a line going halfway down the middle. Each shift creates a new line and you only move halfway to your next line ok — Umonsarmon

Bear with me but I still think there's a problem.

1. For reals less than base-ten 1 you'll get something like 0.xyz...

2. Suppose you take base-ten 0.5 whose binary representation is 0.1

3. You will have to take the "0" in the 0.1 into account or else there's no way you can get a unique fractional counterpart for 1 and 0.1 because if you ignore the "0" you'll be left with 1 for both numbers 1 and 0.1 and that maps to 1/2 to the right in your system.

4. If you take the "0" into account for binary numbers like 0.xyz... then how would you map the numbers 10base2 and -10base2?

5. So 0.1base2 would be 1/2 to the left and then 1/2 of 1/2 to the right giving us 1/2 - 1/4 = 1/4

5. Where would base-ten +2 i.e. 10base2 and base-ten -2 i.e. -10base 2 maps to?

10base2 would be 1/2 to the right and 1/2 of 1/2 to the left i.e. 1/2 - 1/4 = 1/4

-10base2 would be just as above but with a negative sign i.e. -1/4

As you can see three numbers (+2, -2, and 0.5 all base 10) are mapped onto only 2 points. There is no bijection.

His sum, the one he claims can only converge to a rational number, is something like this:

$1/2+\sum_{i=2}^{\infty}(-1)^{k_i}(\frac{1}{2})^i$, where $k_i \in \{0, 1\}$

ETA: Corrected the formula. Thanks

It can be easily shown that the series is convergent by Cauchy's criterion (yes, I just looked up the name - hey, I am three decades out of practice, you guys should be doing this :)). I suspect that it is also order-invariant (if that's the right term), but I won't attempt a proof.

The series can converge to any real number in the interval [0, 1]. There is a simple root-finding numerical method - interval halving, or bisection method - that can be used to demonstrate this. Take the function f(x) = x - r, where r is any real number between 0 and 1. Finding its root in the interval [0, 1] using the interval halving method will produce a series of the above form.

It can be easily shown that the series is convergent by Cauchy's criterion (yes, I just looked up the name - hey, I am three decades out of practice, you guys should be doing this :)) — SophistiCat

Hmmm. The series goes complex for some $k_i$, if you meant the interval of rationals $(0,1)$, I guess you mean $k_i$ is either $0$ or $1$ for all $k_i$. But that makes sense!

The series can converge to any real number in the interval [0, 1]. — SophistiCat

So the procedure says:

"For all input sequence k, the series will converge to a real number in [0,1]"

Now the questions are:

"For all r in the reals, there exists an infinite sequence of inputs such that the series will converge to r?" (this would show the procedure is surjective)

Let's assume that's true. Then the surjection is from infinite binary sequences to infinite binary sequences... Which is unsurprising. It's a surjection from the unit interval of reals to the unit interval of reals, rather than from the unit interval of rationals to the unit interval of reals.

ETA: Corrected the formula:

$\frac{1}{2}+\sum_{i=2}^{\infty}(-1)^{k_i}(\frac{1}{2})^i$, where $k_i$ is 0 or 1

We should be able to prove a stronger claim that a series composed of only positive fractions can converge to any real number between 0 and 1:

$\sum_{i=1}^{\infty}g_i(\frac{1}{2})^i$, where $g_i$ is 0 or 1

We can demonstrate this by a variant of the interval halving method. Let r be a real number between 0 and 1. If r < 1/2, then g₁ = 0, otherwise g₁ = 1. Take the next bracketing interval - [0, 1/2) or [1/2, 1] - and repeat the procedure to find the next g_i. Since each consecutive bracketing interval is half as wide as the previous one, then for any ϵ we can find n such that

$\sum_{i=1}^{n}g_i(\frac{1}{2})^i \in (r-\epsilon, r+\epsilon)$

(I beg your indulgence. It pleases me that I can still solve an elementary calculus problem decades after I took the class :) This thread should probably be moved, since it doesn't really contain any philosophy.)