1) X = .99999... 2) 10X = 9.99999... Subtract 1) from 2) 9X = 9 and X =1. So .9999.... = 1 and all is good. 1) Q = 1 + 2 + 4 + 8......... 2) Q = 1 + 2(1 + 2 + 4 + 8.........) Group terms 3) Q = 1 + 2Q Substitute 1) in 2) Solve for Q: 4) Q = -1 and then -1 = Q = 1 + 2 + 4 + 8......... How can -1 = 1 + 2 + 4 + 8......... ????

Because an infinite geometric series like Q does not converge unless the absolute value of the common ratio is less than 1. So Q does not exist in the mathematical sense, and its value is infinity.

You ruined all the fun. Not rigorous way. If 1) Q = infinity 2) Q + 1 = infinity Subtract 1) from 2) then 1 = 0 Q (infinity) is not a specific number like 400 for example or like you say does not converge to a specific number.

Figuring out your bank account is not advanced calculus, but rather simple addition and subtraction. That's 6 grade stuff, not university advanced math. Besides, the banks do it for you, and figure in their percentage while doing so....

Most of use won't use calculus in our daily jobs, but then again do we really use anything we learned in grade school/college?

Maybe you don't use math in the way you do it on a math test? But having an understanding of math seems to matter. I read of a study that showed that people who could understand ideas like compound interest retired with much more $$ than did people who were ignorant of them.

You pretty much only use Calculus when looking at graphs and want to figure out the amount of something under the curve or above the curve. After 3 years of undergrad research in Chemistry I can honestly say I've used Calculus a handful of times for my research papers. It's kind of a sad thing to admit......

Talker, there is some real confusion with what you have 'assumed' not with maths as such. I would assume some lecturer has given these to you as a teaser, and its worked

What's wrong with the following? Let "L" be the largest real number. then surely 1) L >= 1 Since no real number can be larger than L, L >= L^2 Divide each side by L, and 2) 1 >= L Combining 1) and 2)...1>= L >= 1, so L must = 1 The largest real number is 1. What's going on here?

I think so. Plato's academy didn't want those ignorant of geometry. Descartes, Leibniz, and Bertrand Russell are famous for contributions to mathematics and philosophy.

I think it is more like saying that 1 is the largest number, which is not true. It's clear the conclusion is wrong, but all the ideas leading up to that conclusion seem true. So, is logic unreliable?

What's wrong with this is the step where you have L>=L^2 You cannot both say that L=L^2 and L^2/L=L simply because if L is the largest real number L^2 (which we should then treat as infinity^2)=L so therefore this step should read that L>=L

First, I just want to be sure that there is no misunderstanding about ">=", which is "greater than or equal to", on this keyboard I cannot do the conventional symbol. If I postulate a "largest real number", then it must have the property of being greater than its own square. Mathematicians think such an assumption is nonsense. But in philosophy, this kind of assumption is made in the ontological argument for god. I'm surprised philosophers don't dismiss it as well. i.e. the ontological argument goes something like this. Imagine a most perfect being. Does this being exist? Yes, it must have the attribute of existence, otherwise we've not imagine the most perfect being. Hence, god exists. p.s. Seems the arguments flaw is the assumption that a largest real number exists. Because such a number would have to be bigger than it's own square.

The technological world we thrive in today is thanks to the scientists and engineers who use math everyday

Yes, it takes for granted that L exists, just like the other argument assumed that 1 + 2 + 4 + 8...sums to a finite number. Very true!

Eddy, I just realized that there seems to be no direct way to generate the next prime number, given the previous one, except through various "sieve" schemes which actually just list possible candidates and eliminate the composite ones. In other words, given 7, there seems to be no direct way to find that 11 is the next prime number. Has it been proven to be impossible? Not easy to find the answer from Google. Such a simple problem and no known solution!

This was discussed in "An Excursion Through Number Theory". We can't get a formula to generate primes because they're the numbers that haven't fit into any pattern yet. Something like that. I don't know if it has been proven. I'm at work now, but I'll look at that book when I get home. It's an inexpensive Dover book. I think you'd like it.

These days I limit myself only to material which helps me teach my son. I will be free in a year or so once he is in college. Let me know if there is a proof that it cannot be done.

There is a proof that it cannot be done. On page 38 it says, "...they produce values of Y divisible by p, contradicting the hypothesis. Hence, no such prime producing polynomial exists.

Here's a mindblowing fact. There are infinitely many natural numbers, and infinitely many real numbers, right? Well it turns out that the infinity of the real numbers is larger than the infinity of the natural numbers! To be mathematically precise, what this means is that there is no function from N --> R that is bijective. In other words, no one-to-one correspondence. But get this. It just so happens that there as many rational numbers as natural numbers, and in fact, just as many integers as natural numbers! In mathematics, we call this the cardinality of a set. For finite sets, the cardinality is simply the number of elements but for infinite sets it gets more interesting, and what cantor showed is that |N| < |R|. It has long been conjectured that there is no set S, such that |N| < |S| < |R|. This was known as the continuum hypothesis. However, it won't ever be settled. Kurt Godel and Paul Cohen proved that the continuum hypothesis cannot be proven from the axioms of modern mathematics.

And Cantor's diagonalization proof is ridiculously simple! I was quite surprised by how easy it is. Do you know about infinities higher than R? I think there is one for the number of curves in 2D space? Probably 2 more infinities higher than R?

There are infinitely many infinities higher than R. Take any set S. The set of all subsets of S is known as the power set of S (denoted P(S)), and it will always be the case that |S| < |P(S)|. For finite sets, this is obvious, since if S has n elements P(S) will have 2^n elements, but this extends to infinite sets as well. If interested, here's the proof. Proof: It suffices to show that no function f: S --> P(S) can be surjective. Let A be the set of all elements x in S such that x is not in f(x). Then if x is in A, x is not in f(x), and so f(x) \= A. If x is not in A, then x is in f(x) \= A. Therefore, no element can get mapped to A, so f is not surjective.

What about the number of points in a plane? Is that infinity the same as the infinity of R because (x,y) both come from R? What about the infinity of curves you can draw in a plane? Is it the same one as R?

If your ... term is unlimited, Q = infinite and everything after that are completely meaning less. If not, then you can't go from 1 to 2.

Logic is fine, the way you define and think/treat infinite is not. You should not try to mix infinite and real number.

You can think of a curve as a subset of the plane, but not all subsets of the plane is a curve. I admit, I can't quite prove the number of curves is |R|, but I'm rather certain of it.

Just as I can't prove that a large number of snakes on a plane = academy award, but I'm rather certain of it.