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Are there any proofs about Real numbers that have shorter equivalent proofs going through Complex numbers?

Are there proofs about Integers going through Reals, with longer equivalent proofs using pure Integers?

mcandre
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    Somewhat in that direction: The "hydra problem" would be a theorem about the arithmetic of $\Bbb N$ that is proved by using the arithmetic of the full class of ordinals. – Hagen von Eitzen Jul 29 '16 at 20:54
  • I cannot give you any concrete example right now, however, I'm sure there is something in areas related to 1) factorization of polynomials and fundamental theorem of algebra 2) linear programming with totally unimodular constraint matrices (i.e., these are sure to have some optimal solution which is integral). – dtldarek Jul 29 '16 at 20:56
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    @dtldarek Right, it is certainly cleaner to show factorization into linear factors over $\Bbb C$ than factorization into linear and/or quadratic factors over $\Bbb R$. – Hagen von Eitzen Jul 29 '16 at 20:57
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    A proof of the Prime Number Theorem using elementary means has been given. It is long. Other proofs, including the first one, use analytic methods (calculus). .... Some results on sequences and on finitary combinatorics (e.g. Goodstein's Theorem, and the Paris-Harrington Theorem) imply the consistency of Peano Arithmetic (PA) but can be proven , assuming the existence of an infinite set – DanielWainfleet Jul 29 '16 at 21:41
  • @user254665 It's actually not true that results like Paris-Harrington can be proved "assuming the existence of an infinite set" - e.g. Paris-Harrington is still unprovable in the theory $ACA_0$, which does prove that the set of natural numbers exists. The issue is much more subtle than that - it matters, essentially, what ordinals (or notations for ordinals, rather) the theory in question can prove are well-founded. (Look up "proof-theoretic ordinal.") – Noah Schweber Jul 29 '16 at 22:33

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Many many interesting definite integrals of functions whose indefinite integrals have no closed form are derived using contour integration in the complex plane. Some of these can, however, be derived by far messier work in the real line.

Almost everything involving computational complexity in computer science ends up using the logarithm, which can be defined for only integer arguments if you choose to do so, but whose properties, as a function on the reals, are of great interest and use in complexity analysis.

John Hughes
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    I've not seen an evaluation of $\int_0^{\infty} (1+x^4)^{-1};dx$ done within $\mathbb R. $ But is is a simple contour integration exercise in $\mathbb C.$ – DanielWainfleet Jul 29 '16 at 21:44
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    @user254665 That integral can be solved a number of ways that does not rely on complex analysis. One straightforward way uses partial fraction expansion. In fact, in THIS ANSWER, I developed a closed form solution to the indefinite integral $$I(n)=\int \frac{1}{1+x^n},dx$$for integer $n$. – Mark Viola Jul 29 '16 at 23:22
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Another nice one is in working out which integers may be expressed as the sum of two square integers. Dedekind's proof of the sum-of-two-squares theorem relies on factorising $x^2 + 1$ as $(x+i)(x-i)$.

Patrick Stevens
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    This is a nice example: instead of just normal real or complex values it uses complex numbers with integral coordinates :D – dtldarek Aug 02 '16 at 06:53
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Some time ago I had answered a nice simple question that turned out to be a good example for the Integers/Reals part (I couldn't find simpler discrete proof).

Problem:

Let $G = (U \uplus V, E)$ be a connected bipartite graph such that $n_U = |U| < |V| = n_V$. Then there exists an edge $\{u,v\} \in E$ such that $\deg(u) > \deg(v)$.

Discrete proof:

Sort the vertices such that \begin{align} \deg(u_1) &\geq \deg(u_2) \geq &\ldots &&\geq \deg(u_{n_U}), \\ \deg(v_1) &\geq \deg(v_2) \geq &\ldots &&\geq \deg(v_{n_V}). \end{align}

Observe that because there are no isolated vertices in $G$ we have $\deg(v_{n_V}) \geq 1$, thus $$\sum_{i=1}^{n_U}\deg(u_i)-\deg(v_i) > \sum_{i=1}^{n_U}\deg(u_i)-\sum_{i=1}^{n_V}\deg(v_i) = 0.$$ Let $k$ be the smallest natural number with this property, that is, $$k = \min\Bigg\{k \in \mathbb{N} \ \Bigg|\ \sum_{i=1}^{k}\deg(u_i)-\deg(v_i) > 0\Bigg\}.$$

It follows that

  1. $\deg(u_k) > \deg(v_k)$, otherwise $k$ would not be the smallest number with the above property.
  2. $N\big(\{u_1,\ldots,u_k\}\big) \not\subseteq\big\{v_1,\ldots,v_k\big\}$, because the sum of degrees in $\{v_1,\ldots,v_k\big\}$ is too small to exhaust all the degrees of $\{u_1,\ldots,u_k\big\}$.
  3. There exists an edge $\{u_i,v_j\} \in E$ such that $i \leq k \leq j$.
  4. $\deg(u_i) \geq \deg(u_k) > \deg(v_k) \geq \deg(v_j)$.

Non-discrete proof:

Let $$f\big(\{u,v\}\big) = \frac{1}{\deg(u)} - \frac{1}{\deg(v)}.$$ This is a proper definition, because there are no isolated vertices in $G$. Now observe that $$\sum_{e \in E}f(e) = \sum_{u \in U}\deg(u)\cdot\frac{1}{\deg(u)} - \sum_{v \in V}\deg(v)\cdot\frac{1}{\deg(v)} = |U|-|V| < 0.$$

Therefore, there exists an edge $\{u,v\}$ such that $f\big(\{u,v\}\big) < 0$, but that means $\deg(u) > \deg(v)$.

I hope this helps $\ddot\smile$

Community
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dtldarek
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A nice one is: determination of the radius of convergence of a real power series by analyzing the singularities of the complex function defined by the series.

And more generally: analyzing the asymptotic properties of a sequence $a_n$ by examining the complex singularities of the series $\sum a_n z^n$, the so-called "generating function".

GEdgar
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