I just can't see how leaving out one real number or leaving out a set of 1's and 0's (which ever version of the argument you prefer) demonstrates that the reals tally to a larger infinity.

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I just can't see how leaving out one real number or leaving out a set of 1's and 0's (which ever version of the argument you prefer) demonstrates that the reals tally to a larger infinity.

The point of the proof is that there exist __no__ orderings of the real numbers that permit a bijection to the natural numbers. Such orderings do exist for the rational numbers and even for the algebraic numbers.

The demonstration is that whatever list of reals you have, there is__at least__ one real number that is not in that list. In fact there are uncountably many real numbers that are not in the list, not just one.

The demonstration is that whatever list of reals you have, there is

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Cantor's proof does something different. He doesn't show that there is a mapping from the naturals to the reals which is not a surjection. He shows that

But I can show the exact same thing when listing the rationals. I can use the same kind of rule so that at least one rational is not listed.The point of the proof is that there existnoorderings of the real numbers that permit a bijection to the natural numbers. Such orderings do exist for the rational numbers and even for the algebraic numbers.

The demonstration is that whatever list of reals you have, there isat leastone real number that is not in that list. In fact there are uncountably many real numbers that are not in the list, not just one.

Then do that here, so folks can analyse your work and give feedback that would be accurate in terms of addressing any misunderstandings you may have.But I can show the exact same thing when listing the rationals. I can use the same kind of rule so that at least one rational is not listed.

Here's a cool little way to fit all the rationals inside the natural numbers, with plenty of natural numbers left over.But I can show the exact same thing when listing the rationals. I can use the same kind of rule so that at least one rational is not listed.

I'll do this first for the

The

Now given a fraction $\frac{n}{m}$, I can calculate the natural number $2^n \times 3^m$, which is some natural number.

By the Fundamental Theorem of Arithmetic (FTA), which says that every natural number has a unique factorization into prime powers, we know that no combination of $m$ and $n$ can possibly calculate out to the same number as any other one.

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

So the function $f(n, m) = 2^n \times 3^m$ is one-to-one, or injective, meaning that different pairs go to different natural numbers. It maps

But there are plenty of natural numbers that don't get hit by $f$. For example there's no way to hit $5$. So we can map

Another way to think about this is that perhaps I have to send a secret spy message of vital importance consisting of two natural numbers. But I am only allowed to send out one number, not two. Maybe the spies on the other side can't detect one number, but if I send two they'll be able to capture me.

So I encode my two numbers into one using the above trick; and the receiver of my message simply calculates the prime power factorization of my number and recovers the original pair.

So we see that if we want, we can make the set of rational numbers appear no larger than the set of whole numbers. We can map the rationals into a proper subset of the naturals.

This is a typical cardinality argument. Intuition isn't always reliable. The point in this case is that although SOME mappings from the rationals to the naturals leave out some rationals; other mappings exhaust all the rationals without using up all the naturals!

In fact that furnishes a proof that the rationals and the naturals have the same cardinality; that there are injections going both ways.

But as others have mentioned, there is no injective mapping from the reals to the naturals. Every attempt will result in reals left over. The injection only goes one way.

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You can't show that forBut I can show the exact same thing when listing the rationals. I can use the same kind of rule so that at least one rational is not listed.

You may be trying to say that exactly the same construction works. The problem with this though is that the new number that we construct is not guaranteed to be a rational number. This is where the argument fails for the rationals, as it must because we can demonstrate orderings that suffice.

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Thanks, but I just watched a video from a university lecture, and a student had a similar issue. The professor said that reals can be infinitely long decimals. So I suppose you could never exhaust an irrational like pi with n still as a natural number.You can't show that foreverylisting of rationals.

You may be trying to say that exactly the same construction works. The problem with this though is that the new number that we construct is not guaranteed to be a rational number. This is where the argument fails for the rationals, as it must because we can demonstrate orderings that suffice.

Is that what the argument is essentially about?

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But you could guarantee it to be a rational depending on the rule, right?