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When do the reals start making sense (beyond hand-waving explanations)?

opencl 7y ago
To me at least it was when I took a real analysis course. Very difficult material but once it all 'clicked' I felt like it gave a lot of insight into the fundamentals of math and as a bonus finally offered a rigorous non-hand-wavey explanation of the fundamental theorem of calculus.
stephencanon 7y ago
Most people who say the real numbers make sense doesn't know what they're talking about. The real numbers are pretty bizarre no matter how much mathematics you've studied. You can learn to prove things about them rigorously, and memorize a lot of properties, but very few people will ever develop an accurate intuition for them.
- selimthegrim 7y ago
  There’s a famous story of a seminar on p-adic numbers at Princeton where Harish-Chandra was in attendance and one of the audience brings up some intuition from ordinary primes and the speaker replies with a joke about the degenerate case of the real numbers.
ska 7y ago
Part of any good introduction to real analysis is going to be walking you through a bunch of this stuff until you first understand how much you had glossed over the details, then than the details are strange, and finally that you have a working understanding or at least comfort with techniques to reason about this. Is that the same as making sense? Well to paraphrase a physicist, you don't so much understand it as get used to it. One of the fun things about teaching this material is that usually students have been through years of calculus prior to it, and you get to watch the moment when they realize all of this stuff has been "hiding in plain sight".
- hyperpallium 7y ago
  John Von Neumann https://math.stackexchange.com/questions/11267/what-are-some-interpretations-of-von-neumanns-quote https://math.stackexchange.com/questions/11267/what-are-some... Does the followong seem right to you? Intuition often comes from relating a new thing to something known. So maths, as an abstraction of reality, initially has many sources of intuition. Later maths never has exactly the same patterns as earlier maths (it's already abstracted; so same patterns would be the same thing, though it cam build-on). Eventually, it doesn't relate to anything known, and you have to create that familiarity from scratch. That's hard... but if some stuff became known in the first place, why not this too? (One counter is that the other stuff was instinctively known, e.g. 3D space, or at least our minds are pre-shaped to know it, e.g. language). Assuming math is open-ended, there'll always be new stuff that doesn't relate.
bob457 7y ago
I took two semester of graduate level applied analysis. The reals still don’t make sense to me. Those were definitely the hardest classes I’ve taken.
klyrs 7y ago
By the time you can grok the lesbegue measure? Edit: A great way to unlearn bad intuition is through the study of counterexamples. A good starting point might be chapter 1 of Counterexamples in Analysis [1]. [1]https://pdfs.semanticscholar.org/a4e7/eb352e4c44bf75d8fabaf7594494151cb322.pdf https://pdfs.semanticscholar.org/a4e7/eb352e4c44bf75d8fabaf7...
btilly 7y ago
The reals click for some people, and fail to click for others. What can I say? For me it was once I really understood what a Cauchy sequence of rational numbers is, and how that is a real number. Let's see if I can explain that. In math we have the following kind of construction in lots of places. We take some simple system, we construct some way of representing things from a more complex relationship. And then define some sort of equivalence. The new This is a mouthful but you've seen it. Take the construction of the rational numbers from the integers. A rational number is just a pair of integers (n, m) with the second one not zero. It represents n/m. However there is an equivalence, 1/2 is the same number is 2/4. The equivalence is that (n, m) = (n', m') if and only if n * m' = n' * m. You finish by defining operations as (n, m) + (n', m') = (n * m' + n' * m, m * m') and (n, m) * (n', m') = (n * n', m * m'). This looks like a mouthful, but it is exactly the rule that you're used to. So we've seen this kind of construction before. (You do the same when constructing the integers from the natural numbers.) So constructing the reals from the rationals is done as follows. Intuitively a real is a sequence of rationals that is converging. And two sequences of rationals are equivalent if they should converge to the same thing. Where "converging" means that you have a sequence of rationals (x_1, x_2, x_3, ...) such that if we pick n, m "big enough", then x_n - x_m will be as close to 0 as we want. Or in usual Calculus notation, for every epsilon > 0 there is an N such that for every n and m both bigger than N, abs(x_n - x_m) < epsilon. And (x_1, x_2, x_3, ...) should "converge to the same thing" as (y_1, y_2, y_3, ...) if (x_1 - y_1, x_2 - y_2, x_3 - y_3, ...) converges to 0. Or in usual Calculus notation, for every epsilon > 0 there is an N such that for every n bigger than N, abs(x_n - y_n) < epsilon. You define operations pairwise. So (x_1, x_2, x_3, ...) + (y_1, y_2, y_3, ...) = (x_1 + y_1, x_2 + y_2, x_3 + y_3, ...). Here is a sanity check. If you have a decimal representation, that gives us a sequence of rationals converging to that real, (3, 3.1, 3.14, 3.141, ....). Switch from base 10 to base 2, and you get a different sequence, but it is the same real. And the old chestnut, 1 = 0.99999... repeating is easy to verify. And now work your way through the following axioms: The algebraic axioms are easy. 1. There is a well-defined binary operation + such that x+y is always defined. 2. + is commutative, so x+y = y+x. 3. + is associative, so (x+y)+z = x+(y+z). 4. There is an additive identity 0 such that x+0 = x for all x. 5. Every x has an additive inverse called -x such that x + (-x) = 0 6. There is another binary operation called . 7. is commutative, x * y = y * x 8. * is associative, (x * y) * z = x * (y * z). 9. The distributive property holds. x * (y + z) = (x * y) + (x * z). 10. There is a multiplicative identity 1 different from 0. 11. Every x other than 0 has a multiplicative inverse 1/x such that x * (1/x) = 1. And now the order axiom. 12. Every number is exactly one of positive, negative or 0. Or, more formally, there is a set P closed under addition and multiplication such that for all x, exactly one of three things is true: x is 0, x is in P, or -x is in P. And then the tricky one. Completeness. 13. If X is a non-empty set of reals with an upper bound, it has a least upper bound. (For example the set of x such that x^2 - 2 < 0 is non-empty, it has an upper bound, and therefore it has a least upper bound. Which happens to be sqrt(2).) To see that the order axiom holds, let x_1 be a rational number below the value of something in X, and y_1 be a rational number above an upper bound. And now we construct two sequences as follows. At each step if (x_n + y_n) / 2 is an upper bound, then x_(n+1) = x_n and y_(n+1) = (x_n + y_n) / 2. Else x_(n+1) = (x_n + y_n) / 2 and y_(n+1) = y_n. We can prove three things. 1. (y_1, y_2, y_3, ...) converges to an upper bound. 2. No upper bound can be below what (x_1, x_2, x_3, ...) converges to. 3. Both sequences are equivalent, they represent the same real. The conclusion is that that real has to be the least upper bound. If you can really get that, then congratulations! You understand the reals!
SilasX 7y ago
When you limit yourself to the computable reals.

To me at least it was when I took a real analysis course. Very difficult material but once it all 'clicked' I felt like it gave a lot of insight into the fundamentals of math and as a bonus finally offered a rigorous non-hand-wavey explanation of the fundamental theorem of calculus.

stephencanon 7y ago

Most people who say the real numbers make sense doesn't know what they're talking about. The real numbers are pretty bizarre no matter how much mathematics you've studied. You can learn to prove things about them rigorously, and memorize a lot of properties, but very few people will ever develop an accurate intuition for them.

selimthegrim 7y ago
There’s a famous story of a seminar on p-adic numbers at Princeton where Harish-Chandra was in attendance and one of the audience brings up some intuition from ordinary primes and the speaker replies with a joke about the degenerate case of the real numbers.

ska 7y ago

Part of any good introduction to real analysis is going to be walking you through a bunch of this stuff until you first understand how much you had glossed over the details, then than the details are strange, and finally that you have a working understanding or at least comfort with techniques to reason about this. Is that the same as making sense? Well to paraphrase a physicist, you don't so much understand it as get used to it. One of the fun things about teaching this material is that usually students have been through years of calculus prior to it, and you get to watch the moment when they realize all of this stuff has been "hiding in plain sight".

hyperpallium 7y ago
John Von Neumann https://math.stackexchange.com/questions/11267/what-are-some-interpretations-of-von-neumanns-quote https://math.stackexchange.com/questions/11267/what-are-some... Does the followong seem right to you? Intuition often comes from relating a new thing to something known. So maths, as an abstraction of reality, initially has many sources of intuition. Later maths never has exactly the same patterns as earlier maths (it's already abstracted; so same patterns would be the same thing, though it cam build-on). Eventually, it doesn't relate to anything known, and you have to create that familiarity from scratch. That's hard... but if some stuff became known in the first place, why not this too? (One counter is that the other stuff was instinctively known, e.g. 3D space, or at least our minds are pre-shaped to know it, e.g. language). Assuming math is open-ended, there'll always be new stuff that doesn't relate.

bob457 7y ago

I took two semester of graduate level applied analysis. The reals still don’t make sense to me. Those were definitely the hardest classes I’ve taken.

klyrs 7y ago

By the time you can grok the lesbegue measure? Edit: A great way to unlearn bad intuition is through the study of counterexamples. A good starting point might be chapter 1 of Counterexamples in Analysis [1]. [1]https://pdfs.semanticscholar.org/a4e7/eb352e4c44bf75d8fabaf7594494151cb322.pdf https://pdfs.semanticscholar.org/a4e7/eb352e4c44bf75d8fabaf7...

btilly 7y ago

The reals click for some people, and fail to click for others. What can I say? For me it was once I really understood what a Cauchy sequence of rational numbers is, and how that is a real number. Let's see if I can explain that. In math we have the following kind of construction in lots of places. We take some simple system, we construct some way of representing things from a more complex relationship. And then define some sort of equivalence. The new This is a mouthful but you've seen it. Take the construction of the rational numbers from the integers. A rational number is just a pair of integers (n, m) with the second one not zero. It represents n/m. However there is an equivalence, 1/2 is the same number is 2/4. The equivalence is that (n, m) = (n', m') if and only if n * m' = n' * m. You finish by defining operations as (n, m) + (n', m') = (n * m' + n' * m, m * m') and (n, m) * (n', m') = (n * n', m * m'). This looks like a mouthful, but it is exactly the rule that you're used to. So we've seen this kind of construction before. (You do the same when constructing the integers from the natural numbers.) So constructing the reals from the rationals is done as follows. Intuitively a real is a sequence of rationals that is converging. And two sequences of rationals are equivalent if they should converge to the same thing. Where "converging" means that you have a sequence of rationals (x_1, x_2, x_3, ...) such that if we pick n, m "big enough", then x_n - x_m will be as close to 0 as we want. Or in usual Calculus notation, for every epsilon > 0 there is an N such that for every n and m both bigger than N, abs(x_n - x_m) < epsilon. And (x_1, x_2, x_3, ...) should "converge to the same thing" as (y_1, y_2, y_3, ...) if (x_1 - y_1, x_2 - y_2, x_3 - y_3, ...) converges to 0. Or in usual Calculus notation, for every epsilon > 0 there is an N such that for every n bigger than N, abs(x_n - y_n) < epsilon. You define operations pairwise. So (x_1, x_2, x_3, ...) + (y_1, y_2, y_3, ...) = (x_1 + y_1, x_2 + y_2, x_3 + y_3, ...). Here is a sanity check. If you have a decimal representation, that gives us a sequence of rationals converging to that real, (3, 3.1, 3.14, 3.141, ....). Switch from base 10 to base 2, and you get a different sequence, but it is the same real. And the old chestnut, 1 = 0.99999... repeating is easy to verify. And now work your way through the following axioms: The algebraic axioms are easy. 1. There is a well-defined binary operation + such that x+y is always defined. 2. + is commutative, so x+y = y+x. 3. + is associative, so (x+y)+z = x+(y+z). 4. There is an additive identity 0 such that x+0 = x for all x. 5. Every x has an additive inverse called -x such that x + (-x) = 0 6. There is another binary operation called . 7. is commutative, x * y = y * x 8. * is associative, (x * y) * z = x * (y * z). 9. The distributive property holds. x * (y + z) = (x * y) + (x * z). 10. There is a multiplicative identity 1 different from 0. 11. Every x other than 0 has a multiplicative inverse 1/x such that x * (1/x) = 1. And now the order axiom. 12. Every number is exactly one of positive, negative or 0. Or, more formally, there is a set P closed under addition and multiplication such that for all x, exactly one of three things is true: x is 0, x is in P, or -x is in P. And then the tricky one. Completeness. 13. If X is a non-empty set of reals with an upper bound, it has a least upper bound. (For example the set of x such that x^2 - 2 < 0 is non-empty, it has an upper bound, and therefore it has a least upper bound. Which happens to be sqrt(2).) To see that the order axiom holds, let x_1 be a rational number below the value of something in X, and y_1 be a rational number above an upper bound. And now we construct two sequences as follows. At each step if (x_n + y_n) / 2 is an upper bound, then x_(n+1) = x_n and y_(n+1) = (x_n + y_n) / 2. Else x_(n+1) = (x_n + y_n) / 2 and y_(n+1) = y_n. We can prove three things. 1. (y_1, y_2, y_3, ...) converges to an upper bound. 2. No upper bound can be below what (x_1, x_2, x_3, ...) converges to. 3. Both sequences are equivalent, they represent the same real. The conclusion is that that real has to be the least upper bound. If you can really get that, then congratulations! You understand the reals!

SilasX 7y ago

When you limit yourself to the computable reals.