4 ms·

> I suspect a lot of mathematicians prefer the format given because it is more obtuse... No, because as you say, defining what a rational number is seems prett

by jordigh 11y ago

> I suspect a lot of mathematicians prefer the format given because it is more obtuse... No, because as you say, defining what a rational number is seems pretty pointless here, as why would you be reading the proof of something whose definition you don't even know? And you also want an explanation of what a proof by contradiction is, which also seems to be way too elementary. Proof by contradiction is one of the most basic techniques. Spivak's version of this proof explains it a bit more, but still requires work from the reader. Any proof does. Mathematics cannot be a spectator sport.

Natsu 11y ago
A lot of people struggle at math because they're missing things like that. With today's technology, there's no reason things can't be broken down to arbitrarily small steps, so that people can be filled in on the cracks in the foundations of their understanding of math. The usual reason for not including everything is that it becomes ridiculously laborious. With mathematics being abstraction piled atop abstraction, this is quite reasonable. Just look at how much work it is to prove 2+2=4 -- http://us.metamath.org/mpegif/mmset.html#trivia http://us.metamath.org/mpegif/mmset.html#trivia But, as you can see from the above, there's technology to simplify things. I wonder if someday we'll be able to break things down in a friendly way so that for any piece a student doesn't understand, they can get a proof in terms of things they do understand?
- chx 11y ago
  Erm, the linked metamath page asks what's the longest path one can take. Well, geez, it's long. Proving that 2+2=4 from the usual Peano axioms isn't that hard or long. We define addition by a+0=a,a+S(b)=S(a+b). Now what we mean by a positive integer number is just a shorthand for that number of successive S(). So 2+2=4 is just S(S(0)) + S(S(0)) = S(S(S(S(0)))). To prove this, let's apply the second part of the addition to S(S(0)) + S(S(0)) we get S(S(S(0)) + S(0)). Let's repeat this "move the S() from the second to the outside" and you will get S(S(S(S(0)) + 0)). Now we can use our first half of definition where +0 can be left out: S(S(S(S(0)))). Now, that wasn't that long, was it?
  - lokedhs 11y ago
    I'm not a mathematician, so please bear with me. I'm trying to understand your definition of addition. Surely that can't be enough? At the very least a+b=0 satisfies your formulas (should I use the word axiom there?). What I'm trying to understand is how I am supposed to think in order to prove/disprove my statement above. Disproving something by contradiction is easy enough, if you can find a contradiction but what if you can't? How can you be sure that you've covered all variations? This is something one struggles with as a developer too. There are cases where you think that a fundamental function does the right thing for all inputs but then you discover an edge case where that isn't true. The number of cases where this has been the case even in high profile libraries suggests that I'm not alone in this.
    - gizmo686 11y ago
      What is missing from chx's description is the definition of the natural numbers (and the S (successor) function). The standard definition is: 0 is a natural number S(n) is a natural number, if and only if n is a natural number x=y if and only if S(x)=S(y) for all x, S(x)!=0 These axioms are sufficient to show that the natural numbers form a line. At this point, we can use the definition of addition that chx provided: a+0=a a+S(b)=S(a+b) It is true that, without the restrictions on S, than defining a+b=0 would satisfy these. However, because of the restrictions on S from our definition of the natural numbers, it is impossible for S(a+b)=0.
  - Natsu 11y ago
    In this case, both of the twos are actually complex numbers (2+0i) so they have to construct complex arithmetic from bare axioms.
- semi-extrinsic 11y ago
  In some sense, proofs as presented in a textbook should not hold your hand. The point of learning about proofs is being able to create proofs yourself, even if they are trivial variations on proofs you have seen. Computer analogy: would it be a good idea if someone showing an implementation of a double-linked list explained every little detail about pointers and arrays etc.? However, there are many things we can do syntax-wise to facilitate the understanding of proofs. I remember we experimented with using indentation (on paper) in proofs when I did functional analysis, and that was quite useful for keeping track of "variable scope" etc.
  - Natsu 11y ago
    While it's true that you won't survive long with a lot of 'cracks' in your foundation of understanding, I think there's quite a bit of worth in helping people identify and explore the holes in their own understanding and making this as painless as possible. I understand that we can't trod new ground if nobody is willing to put up with the difficult slogs, but there's no reason to make things difficult for the sake of difficulty. There are plenty of challenging exercises one can do simply by learning to apply a newfound understanding without having to leap from perch to perch because nobody would explain how to bridge a certain gap. I think that things are trending in this direction already, actually.
- antognini 11y ago
  I'm reminded of the textbook Axiomatic Set Theory by Suppes, which has on page 137 a proof that 3 + 1 = 4.

by jordigh 11y ago

Natsu 11y ago

A lot of people struggle at math because they're missing things like that. With today's technology, there's no reason things can't be broken down to arbitrarily small steps, so that people can be filled in on the cracks in the foundations of their understanding of math. The usual reason for not including everything is that it becomes ridiculously laborious. With mathematics being abstraction piled atop abstraction, this is quite reasonable. Just look at how much work it is to prove 2+2=4 -- http://us.metamath.org/mpegif/mmset.html#trivia http://us.metamath.org/mpegif/mmset.html#trivia But, as you can see from the above, there's technology to simplify things. I wonder if someday we'll be able to break things down in a friendly way so that for any piece a student doesn't understand, they can get a proof in terms of things they do understand?

chx 11y ago
Erm, the linked metamath page asks what's the longest path one can take. Well, geez, it's long. Proving that 2+2=4 from the usual Peano axioms isn't that hard or long. We define addition by a+0=a,a+S(b)=S(a+b). Now what we mean by a positive integer number is just a shorthand for that number of successive S(). So 2+2=4 is just S(S(0)) + S(S(0)) = S(S(S(S(0)))). To prove this, let's apply the second part of the addition to S(S(0)) + S(S(0)) we get S(S(S(0)) + S(0)). Let's repeat this "move the S() from the second to the outside" and you will get S(S(S(S(0)) + 0)). Now we can use our first half of definition where +0 can be left out: S(S(S(S(0)))). Now, that wasn't that long, was it?
- lokedhs 11y ago
  I'm not a mathematician, so please bear with me. I'm trying to understand your definition of addition. Surely that can't be enough? At the very least a+b=0 satisfies your formulas (should I use the word axiom there?). What I'm trying to understand is how I am supposed to think in order to prove/disprove my statement above. Disproving something by contradiction is easy enough, if you can find a contradiction but what if you can't? How can you be sure that you've covered all variations? This is something one struggles with as a developer too. There are cases where you think that a fundamental function does the right thing for all inputs but then you discover an edge case where that isn't true. The number of cases where this has been the case even in high profile libraries suggests that I'm not alone in this.
  - gizmo686 11y ago
    What is missing from chx's description is the definition of the natural numbers (and the S (successor) function). The standard definition is: 0 is a natural number S(n) is a natural number, if and only if n is a natural number x=y if and only if S(x)=S(y) for all x, S(x)!=0 These axioms are sufficient to show that the natural numbers form a line. At this point, we can use the definition of addition that chx provided: a+0=a a+S(b)=S(a+b) It is true that, without the restrictions on S, than defining a+b=0 would satisfy these. However, because of the restrictions on S from our definition of the natural numbers, it is impossible for S(a+b)=0.
- Natsu 11y ago
  In this case, both of the twos are actually complex numbers (2+0i) so they have to construct complex arithmetic from bare axioms.
semi-extrinsic 11y ago
In some sense, proofs as presented in a textbook should not hold your hand. The point of learning about proofs is being able to create proofs yourself, even if they are trivial variations on proofs you have seen. Computer analogy: would it be a good idea if someone showing an implementation of a double-linked list explained every little detail about pointers and arrays etc.? However, there are many things we can do syntax-wise to facilitate the understanding of proofs. I remember we experimented with using indentation (on paper) in proofs when I did functional analysis, and that was quite useful for keeping track of "variable scope" etc.
- Natsu 11y ago
  While it's true that you won't survive long with a lot of 'cracks' in your foundation of understanding, I think there's quite a bit of worth in helping people identify and explore the holes in their own understanding and making this as painless as possible. I understand that we can't trod new ground if nobody is willing to put up with the difficult slogs, but there's no reason to make things difficult for the sake of difficulty. There are plenty of challenging exercises one can do simply by learning to apply a newfound understanding without having to leap from perch to perch because nobody would explain how to bridge a certain gap. I think that things are trending in this direction already, actually.
antognini 11y ago
I'm reminded of the textbook Axiomatic Set Theory by Suppes, which has on page 137 a proof that 3 + 1 = 4.