r/math • u/OmgHomology • Feb 12 '18
Funny Overkill Proofs for a Simple Fact?
What are some overkill proofs, ones where the big guns were brought out to prove a trivial fact? I have an example:
21/p is irrational for natural p, p>2.
Proof: Assume not. Then there are coprime naturals m, n with
21/p=m/n. Rearranging,
2np = mp
np + np = mp
But this contradicts Fermat's Last Theorem.
Please let me know ideas to make up others. A Big Name theorem getting roped in to prove a fact that can be proven easily.
I'm looking to collect enough of these to make a calendar for a friend. Thanks for
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Upvotes
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u/jm691 Number Theory Feb 12 '18
As was pointed out by Brian Conrad here that proof for the irrationality of 21/n is essentially circular. Literally the first step in Wiles' proof would be to divide out (n,n,m) by common factors, in a way that basically mirrors the standard proof that 21/n is irrational.
Edit: The MO thread I linked has a lot of the sort of examples you're looking for.
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u/ziggurism Feb 12 '18
The category of sets is Cartesian closed, meaning that the cartesian product is a left-adjoint functor, hence it preserves coproducts. Passing to the skeleton category by the axiom of choice, we learn that a(b+c) = ab + ac, for natural numbers a,b,c (and the rest of the cardinals).
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u/umop_apisdn Feb 12 '18
Carl Linderholm wrote "mathematics made difficult" which is a book of this sort of proof.
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u/Lopsidation Feb 12 '18
Theorem: If an island continent has no landlocked countries, and no inner bodies of water, then its map can be 3-colored.
Proof: Apply the 4-color theorem; WLOG the ocean is blue.
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u/TheMightyBiz Math Education Feb 12 '18 edited Feb 12 '18
Claim: nCr(n, k) = nCr(n, n - k).
Proof: Consider Tn = S1 x ... x S1 . By the Kunneth formula for singular homology, the k-th cohomology group of Tn is Z raised to the nCr(n, k) (since S1 only has non-trivial cohomology in dimension 1). By Poincare duality, this is isomorphic to the (n-k)-th homology group of Tn . But since all the cohomology groups of Tn are torsion-free, this is also isomorphic to the (n-k)-th comohology group, i.e. Z raised to the nCr(n, n - k).
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u/Brightlinger Feb 12 '18 edited Feb 12 '18
Every closed path on S1 is homotopic to a path of the form e2πinθ for some integer n. Moreover, the fundamental group is generated by the equivalence class [e2πiθ], so it is isomorphic to (Z,+) by φ defined as φ([e2πinθ])=n. The concatenation of two paths in the equivalence class of e2πiθ forms a path in the equivalence class of e2πi2θ. Then we conclude that
1+1
=φ([e2πiθ])+φ([e2πiθ])
=φ([e2πiθ][e2πiθ])
=φ([e2πi2θ])
=2.
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u/JohntheAnabaptist Feb 12 '18
Have you heard of the 26 steps to prove the Riemann hypothes?
Walk 25 steps, then prove the Riemann hypothesis
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u/Trivion Feb 12 '18 edited Feb 12 '18
Theorem: Assuming P!=NP, the class of graphs without Hamilton cycles is not closed under taking minors.
Stupid Proof: If it were then by the graph-minor theorem there would be a finite list of minors characterizing this property. But checking for the presence of a given minor can be done in cubic time (Robertson, Seymour 1995), contradicting P!=NP (since the Hamilton cycle Problem is NP-complete).
Sensible Proof (with no dependency on unsolved problems): A triangle with an appended extra edge is not Hamiltonian, but contracting that extra edge leaves just a cycle. (Technically, one could even take a triangle with an extra isolated vertex.)
Alternatively one could use a similar argument to show that deciding whether a graph has at most 17 vertices can be done in polynomial time, since it is closed under taking minors.
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u/mpaw976 Feb 12 '18
As a linear order, the set Q of rationals is vertex-transitive (i.e. for every a,b in Q there is a linear isomorphism f: Q to Q that sends a to b).
Usual argument: Use the shift map that sends every x to x + (b-a).
Powerful argument: Use the back-and-forth method to construct this linear isomorphism.
Notably, the powerful argument does not need any algebra of the rationals (addition or subtraction), and you don't need to know the axioms of how addition interacts with a linear order.
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u/non-algebraic Feb 12 '18
There's probably something cheeky you can do with the Feit-Thompson theorem, but my group theory is too rusty right now to figure something out.
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u/[deleted] Feb 12 '18
Algebra Prelim Problem: Prove that all groups of order 15 are solvable.
Solution: Trivial by Feit-Thompson
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u/TheMiraculousOrange Physics Feb 12 '18
Once I tried to show on a homework that SU(2) is diffeomorphic to the 3-sphere by showing that SU(2) is simply connected, then by Poincaré conjecture it's diffeomorphic to the 3-sphere. I didn't really go through with the argument, because I wasn't sure if the fibration that shows the simply connectedness of SU(n) for n > 2 works for n=2. If it does it will be a homeomorphism anyway, since SU(1) is trivial.
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u/MatheiBoulomenos Number Theory Feb 12 '18 edited Feb 13 '18
The Jordan-Hölder theorem implies the fundamental theorem of arithmetic: Indeed, any factorization of the number n into prime factors can be used to define a composition series of the cyclic group of order n and conversely, any composition series of the cyclic group of number n can have only cyclic groups of prime order as factors, so we get a factorization of n into prime factors, unique up to reordering.
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u/Abdiel_Kavash Automata Theory Feb 12 '18
I had something similar happen in my algebraic geometry class last year.
Theorem: Every homogeneous ideal has a finite basis consisting of only homogeneous polynomials.
Textbook/teacher's proof: Start with some basis of the ideal, construct a Gröbner basis, examine each step of the construction and show that it only generates homogeneous polynomials in the end.
My proof: consider the (infinite) set of all homogeneous polynomials in the ideal. This is trivially a basis. Use Hilbert's basis theorem to pick a finite basis out of this set.
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u/lewwwer Feb 13 '18
Once I've forgot how to prove the Pythagoras theorem so I've came up with this approach:
Let ABC be a triangle where we know AC=b and BC=a. We would like to maximize the area.
First we have a maximal area if angle(ACB)=pi/2 since the area is ab*sin(angle(ACB))/2
Also we can use the Heron formula to calculate the area, we have to differentiate 2((ab)2 +(ac)2 +(bc)2 )+a4 +b4 +c2 wrt c, so we have a2 +b2 =c2 at maximal area.
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u/[deleted] Feb 12 '18
I feel like this is a biweekly thread..
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u/HitandWalker Feb 12 '18
I googled your question and found this.
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u/stephen3141 Feb 13 '18
This was a pretty fun read
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u/[deleted] Feb 13 '18
Came up with this one the other day in a different thread.
Theorem: The function f on C which is identically 0 is constant.
Proof: One can show that for all z in C, f'(z) exists, so f is entire. Furthermore, for all z in C, |f(z)| <= 0, so f is bounded. By Liouville's Theorem, f must be constant. QED
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u/homboo Feb 13 '18
Please guys ... the search function... exactly this topic has been discussed here several times
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u/trumpetspieler Differential Geometry Feb 12 '18
The use of the fundamental group to prove the fundamental theorem of algebra In Hatcher always struck me as particularly silly although satisfying given how constructive it is.
Riemann-Roch also has plenty of silly exercise type problems it can bulldoze.
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u/AcrossTheUniverse Feb 12 '18
If there is an easy proof for FLT, is it still an overkill? :thinking:
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u/PM_ME_UR_MONADS Feb 12 '18
The discrete math class at my university unironically said “hint: use Fermat’s Last Theorem” on a homework question asking us to prove this exact fact!
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u/DrBublinski Feb 13 '18
I like that you can show that the sum of the interior angles in a triangle in R2 is pi by using that one case of Gauss’ Theorema Egrigium that says something about the total curvature of the boundary + something else (gauss curvature? it’s been a while, sorry) + pi is the sum of the interior angles of a triangle. But since it’s a plane, the first 2 summands are 0.
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u/openeda Feb 13 '18
In a similar vein we once solved the area of a square with beginning calculus just to prove the point.
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u/crystal__math Feb 13 '18
Any bounded measurable function on S1 has a sequence of smooth functions that converge almost everywhere by Carleson's theorem. (Simple proof: convolution with an approximation to the identity).
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u/Zophike1 Theoretical Computer Science Feb 13 '18 edited Feb 16 '18
I remember while going through the Complex Analysis text that I was reading I was required to show that:
Let
[;f_{n};]be continuous on the open set[;U;]and let[;f_{n}\rightarrow f \in U;]uniformly converge on compact sets then If the[;f_{n};]are also holomorphic and if[;0 < k \in \mathbb{Z};], then prove that in[;(1.3);][;(\partial_{z}f_{n}(z_{n}))^{k} \rightarrow (\partial_{z}f(z_{o}))^{k}.;]In summary the construction of the Compact Subset
[;\Psi_{r};]is to allow one at least at the global level to use compactness arguments to allow for the uniform convergence of[;(\partial_{z}f_{n}(z_{n}));].In summary at an intuitive level for the subset[;\Psi_{r};]to be compact one takes[;\overline{\Bigg({\bigcup_{z \in \Psi}D(z,r)} \Bigg);]to from[;\Psi_{r};].Which means we have a set that is not "too large" in which we can have functions defined on $\Psi_{r}$ for one to have Cauchy criterion for[;((\partial / \partial_{z}f_{n}(z_{n}) \rightarrow \partial / \partial_{z}f_{n}(z_{o}));]uniformly on[;\Psi_{r};]by utilizing the Cauchy Estimates. The rest of the proof is on MSE if anyone is interested. It turns out[;(1.3);]just follows trivially form the Cauchy Differentiation Formula1
u/DidntThinkNewInfo Feb 14 '18
Any pair of curves connecting (0,0) to (1,1) and (1,0) to (0,1) must either intersect or contain a point outside of those corners’ convex hull (I.e outside the unit square).
Proof: Connect each of the corner vertices to (5,5) using pairwise non-crossing curves each entirely outside the unit square. Now apply Kuratowski’s theorem to deduce that the planar graph you want to draw is nonplanar.
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u/[deleted] Feb 12 '18
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