“Mathematical Intuition—What Is It?”

The article below is related to a strand or theme of articles I am publishing today on knowledge, famous science, and people.–kas

Mathematical Intuition—What Is It?

by rjlipton

Why is math intuition hard for some areas and less for others?

Enrico Bombieri is one of the world leaders in many areas of mathematics, including number theory, algebraic geometry, and analysis. He has won many awards, and is a Fields Medalist.

Today I want to talk about the notion of intuition in mathematics. I am curious what it is, how to get it, and how to use it.

One story, perhaps an urban legend, is that a senior expert in real analysis was once sent a paper that “proved” a surprising theorem. The expert looked at the proof, and was immediately skeptical. The “theorem” seemed to be too surprising—his intuition based on his great experience, was that the theorem could not be true. Yet even after hours of studying the proof he could not find any mistakes. But his intuition continued to bother him. He finally looked even more carefully, and found the problem. The author of the proof had used a lemma from a famous topology book. He had used the lemma exactly as it was stated in the famous textbook. But there was a typo in the book. Somehow the words “closed” and “open” had been exchanged in the statement of the lemma. This made the lemma false, caused a gap in the proof of the surprising theorem, and left the poor author with a buggy paper.

Proving theorems is not mechanical; proving theorems does require formal manipulation. Yet proving theorems also requires the use of intuition, the ability to see what is reasonable or not, and the ability to put all these together. Blindly using a lemma from even the most famous textbook can be dangerous, as the story shows.

I once lost several months of hard work trying to use a published theorem to solve an open problem. I almost had a proof, but eventually like the story I found a bug in the published result—the result was the sole result of a friend’s Ph.D. thesis. Oh well. This is not an urban legend, I was there, but I will discuss it another time.

For now let’s turn to the discussion of intuition in mathematics.

In mathematics you don’t understand things. You just get used to them. John von Neumann


Number Theory

I think that it is not too hard to have a reasonable intuition in number theory, especially concerning the behavior of prime numbers. A pretty fair approximation is to try the distribution of primes as “random.” That is primes in the range {[N,N+\Delta]} are roughly randomly selected odd numbers with the correct density: the number of primes in such an interval for {\Delta} is about {\Delta/\ln N}. This of course follows from the prime number theorem.

A classic example of this is the conjectured behavior of twin primes. If primes are random, then one would expect that there are about

\displaystyle  C N /\ln^2 N

twin primes in {[1,N]} where {C} is a constant. Godfrey Hardy and John Littlewood made an even more detailed conjecture, which included the value of the constant {C}. They guess that {C} is

\displaystyle  \prod_{p \ge 3} \frac{p(p-2)}{(p-1)^2} \approx 0.6601\dots

Of course there are local properties that need to be added to the model. The number of primes {p} so that {p+2} and {p+4} are all primes is not

\displaystyle  C N /\ln^3 N,

but one. This follows since on of {p}, {p+2}, {p+4} is divisible by {3}: so {p} must be equal to {3}.

I once had a proof that needed only a lemma about the structure of the primes to be complete. It was about a communication lower bound that I was working at the time with Bob Sedgewick. We could not prove the lemma, nor could we find any references to anything like it. So we made an appointment to see the famous number theorist, Enrico Bombieri. He is a member of IAS, and was there at the time. So Bob and I went over to ask him about our lemma.

Bombieri listened very politely, asked a question or two for clarification, and said, “yes that lemma is surely correct.” We were thrilled, since this would make our proof complete. I then asked him for a reference. He looked at me and said:

Of course the lemma about primes is true, but it is completely hopeless to prove it.

He had great intuition about primes, but proving certain results was then and still is today completely beyond anything anyone can do.



My intuition in geometry, especially high dimension, is very poor. I have proved basic geometry theorems that hold in {n}-dimensions, but I have terrible geometric intuition.

I am not sure why, I am not sure what makes geometry so hard for me, but it is very different than prime numbers. There are plenty of geometric questions that I would not have any idea how to conjecture what is true. Perhaps it is just a defect in my abilities, but geometry seems to be orders of magnitude trickier than number theory.

See this for some nice comments on high dimension geometry. It is a quite interesting anonymous blog—it is being written a by a Ph.D. student, I wish him/her well.



My intuition about finite groups is even worse than my geometric intuition. No that is not quite right. In a sense my intuition about groups is really very good. Over the years I have hit upon a rule in thinking about groups. I figured out that if I thought that X was a reasonable theorem that should hold for finite groups, then X was likely to be false.

Of course, this is a bit silly. It is like having a really poor sense of picking stocks—if you were always wrong, then there would be a great strategy. But, somehow I do believe there is something to what I am saying. My intuition is so bad that after a while I just started to guess the opposite of whatever I first thought.

My last post is a perfect example. I have a series of conjectures about solvable groups. The conjecture I listed took an expert, Colin Reid, a few minutes to disprove. He is a group theorist. I should have known better, but my intuition about groups is terrible, yet they may play an important role in our understanding of theory.


Reid’s Proof

Colin Reid posted his proof that {\cal SOLVE} is impossible in the comments section of the last discussion. The parts of his comments using angle brackets were snipped as HTML-style tags, so this is an expanded version fixing the glitches. Ken Regan and I have also replaced his quotient-subgroup notation by homomorphism notation.

Let {G} be a non-trivial solvable group—some say soluble group. The composition series is defined by {G_0 = G}, and for {i \geq 1}, {G_i = [G_{i-1},G_{i-1}] =} the subgroup generated by the commutators {[u,v] = uvu^{-1}v^{-1}} for {u,v \in G_i}. Solvability of {G} means that some {G_i} is the trivial subgroup {\{1\}}, whereupon the series terminates. The important facts are that not only is every {G_i} normal in its predecessor, it is normal in the entire group {G}, and that the immediate quotients {G_{i-1}/G_i} are abelian. This means that we can define a homomorphism {\pi} on all of {G} whose kernel is {G_i}, and for all {c,d \in G_{i-1}}, {\pi(c)} and {\pi(d)} commute. The reason for the latter is that

\displaystyle dc = cdk

where {\pi(d)\pi(c) = \pi(dc) = \pi(cdk) = \pi(c)\pi(d)\pi(k) = \pi(c)\pi(d)}, since {k} is in the kernel {G_i}.

Now recall that {\cal SOLVE} asserted the existence in {G} of elements {a,b,c,d} such that some conjugates {xax^{-1}} of {a} and {yby^{-1}} of {b} are in {\langle c,d \rangle}, where {x,y \in G}, and likewise some conjugates of {c} and {d} are in {\langle a,b \rangle}, with a third condition on the orders of these elements in {G}. The condition is that {o(a)o(b)} be relatively prime to {o(c)o(d)}, and this is what finally prevents the existence of {a,b,c,d}.

For contradiction, suppose a solvable group {G} with such elements exists. Then in the composition series, there is some {i} such that {G_{i-1}} contains all of {a,b,c,d}, but {G_i} does not. By symmetry, without loss of generality, we can suppose {G_i} does not have {a}. Take {a' = xax^{-1}} as above, so {a' \in \langle c,d \rangle}. It does not matter whether {c} or {d} belongs to {G_i}; that both belong to {G_{i-1}} is enough. Since {a'} is a conjugate, {o(a') = o(a)}. Now take the homomorphism {\pi} with {G_i} as kernel, and observe:

  1. {\pi(a') \in \langle \pi(c),\pi(d) \rangle}.
  2. {o(\pi(c))} divides {o(c)}, {o(\pi(d))} divides {o(d)}, and {o(\pi(a'))} divides {o(a') = o(a)}.
  3. Since {c} and {d} are in {G_{i-1}}, {\pi(c)} and {\pi(d)} commute.
  4. Hence {\pi(c)} and {\pi(d)} can generate at most {m = o(\pi(c))o(\pi(d))} different elements.
  5. Put more strongly, the subgroup {\langle \pi(c),\pi(d) \rangle} they generate is also a subgroup of the abelian group generated by {\pi(c)} and {\pi(d)}, which has exactly {m} elements, so by Lagrange’s theorem, the order of {\langle \pi(c),\pi(d) \rangle} divides {m}.
  6. Since {\pi(a')} is in {\langle \pi(c),\pi(d) \rangle}, it follows that {o(\pi(a'))} divides {m}, which divides {o(c)o(d)}.
  7. However, since {o(\pi(a'))} divides {o(a)} which is relatively prime to {o(c)o(d)}, the only way this can happen is {o(\pi(a')) = 1}.
  8. That means {\pi(a') = 1}, so {a' \in G_i} since {G_i} is the kernel of {\pi}.
  9. But {G_i} is normal in {G}, not just in {G_{i-1}}, so {x a' x^{-1}} is in {G_i}. This puts {a} into {G_i}, which yields a contradiction.

To someone with good algebraic intuition this comes trippingly off the tongue, which is why it is a service to communicate in public. So we thank Colin Reid—and we will see if the insight gained works against our more-complicated stratagems of this kind. Colin is currently a grad student at Queen Mary college, University of London.

It may be that deterministically simulating each level of a Boolean circuit simply must bump you one step along a composition series, which in a solvable group {G} is a finite, non-renewable resource.

However, we also have other ideas\dots Perhaps we can get mileage out of choosing different solvable groups {G} for different input sizes {n}. This relates complexity questions to ones about the possible lengths of composition series in groups of certain sizes—although what we know about such sizes is not so promising. But perhaps a randomized simulation can possibly avoid these limitations.


Complexity Theory

I think we have good intuition here, but I have seen many surprises in my career:

  • Linear Programming is in polynomial time.
  • Nondeterministic space is closed under complement.
  • Polynomial size bounded branching programs are powerful.
  • Permanent has random self-reduction.
  • Quantum polynomial time can factor integers.
  • De-randomization of random walks for undirected graphs.
  • Proofs can be checked in constant time.
  • The existence of Zero-Knowledge protocols.
  • {\dots}

I have reasonable intuition, yet all of these were surprising to me. Even some that I worked on and contributed to their development.


Open Problems

Is intuition simply built up by learning more and more about an area? Or is intuition something that is separate from just being an expert in an area? Can you be quite strong in an area and still have weak intuition, or is that impossible?



from: http://rjlipton.wordpress.com/2010/10/01/mathematical-intuition-what-is-it/

About eslkevin

I am a peace educator who has taken time to teach and work in countries such as the USA, Germany, Japan, Nicaragua, Mexico, the UAE, Kuwait, Oman over the past 4 decades.
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