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Converse of Pythagoras' theorem: If the lengths of the sides of a triangle $T$ are $a$, $b$, and $c$, and if $a^2+b^2=c^2$, then the triangle is a right triangle and the side opposite to the right angle is the one whose length is $c$.

Proof: Construct a line segment $XY$ whose length is $a$. Then construct a line segment $YZ$ whose length is $b$ which is perpendicular to $XY$. By construction, the triangle $XYZ$ is a right triangle, and therefore, by Pythagoras' theorem and because we are assuming that $a^2+b^2=c^2$, the length of $XZ$ is equal to $c$. So, the triangle $XYZ$ is similar to the original triangle $T$. Since the triangle $XYZ$ is a right triangle, then so is $T$.

What I find peculiar about this proof is the fact that it uses Pythagoras' theorem in order to prove its converse.

It is not the only situation that I am aware of in which this occurs. For instance, there is a proof of the converse of Ceva's theorem which uses that theorem. But I am not aware of any example outside Euclidean Geometry.

Can anyone provide an example of a theorem of the type $A\implies B$ outside Geometry with a proof which uses the fact that $B\implies A$?

Alex Ortiz
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José Carlos Santos
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  • Related? https://math.stackexchange.com/questions/3572137/if-propositions-p-and-q-are-equivalent-and-p-implies-r-without-using-q – BCLC Nov 23 '20 at 12:56
  • wait how could you possibly use B implies A when we're trying to prove B? Using B implies A suggests to me that B is somehow already true – BCLC Nov 23 '20 at 12:59
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    @JohnSmithKyon Did you actually read my question? It is about using the fact that $B\implies A$ to prove that $A\implies B$. At no moment I mention the possibility of proving $B$ alone. – José Carlos Santos Nov 23 '20 at 13:46
  • I didn't actually read. I just kinda skimmed. I sorta scanned but didn't really analyse. But still the fact that you use B implies A you means ostensibly that you have satisfied the assumptions of 'B implies A', i.e. B. Eh you can just ignore this if you really think that my question is answered if I were to really analyse the post. – BCLC Nov 23 '20 at 13:56
  • Apart from Menezio contribution all answer to this big list are mine :( – nonuser Nov 03 '21 at 17:54
  • @Aqua Indeed! And I have upvoted every one of them. When I first posted the question, I did not expect that it would get so many answers. – José Carlos Santos Nov 03 '21 at 18:06

6 Answers6

15

Say natural number $n$ is good if it can be writen as a sum of two squares.

Theorem: $n$ is good iff $2n$ is good.

Proof: If $n=a^2+b^2$ then $2n = (a+b)^2+(a-b)^2$ and we are done.

Now the converse. Say $2n$ is good. Then, by already proven part, also $4n$ is good, so $$4n = x^2+y^2$$ Since $x,y$ must be both even, we can write $a=x/2$ and $b=y/2$ and we are done.

nonuser
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4

Let $f: A \to B$ be a function and $g: A \to A$, $h: B \to B$ be bijections.

Prove that $f$ is surjective if and only if $h \circ f \circ g$ is surjective.

Proof: Suppose $f$ is surjective .... we get $ h \circ f \circ g$ is surjective.

Now the converse: Suppose $h \circ f \circ g = :e$ is surjective. Since $g^{-1}$ and $h^{-1}$ are surjective then, by already proven part, we have $h^{-1} \circ e \circ g^{-1}=f$ is surjective.

nonuser
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2

Definition: Graph $G$ is perfect if each $H\leq G$ has chromatic number the same as clique number.

Theorem: (László Lovász, 1972) Graph $G$ is perfect iff it complement $\overline{G}$ is perfect.

Proof:

  • Suppose if part is proven.
  • Now let $\overline{G}$ be perfect. Then by already proven part also $\overline{\overline{G}}$ is perfect. But $\overline{\overline{G}} =G$ and we are done.
nonuser
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1

I've just found the following :

Let $K$ an infinite field. $f(x_0,...,x_n)\in K[x_0,...,x_n]$ is homogeneus of degree $d$ $\Longleftrightarrow$ $f(\lambda x_0,...,\lambda x_n) = \lambda^d f(x_0,...,x_n)$ for all $\lambda\in K$.

Proof: $\Rightarrow$) For each monomial of $f(\lambda x_0,...,\lambda x_n)$ you can take out a factor $\lambda^d$. Hence we have the statement (basically is the definition).

$\Leftarrow$) Suppose $f(x_0,...,x_n)=\sum_{i=1}^kf_{j_i}(x_0,...,x_n)$ where $f_{j_i}$ are homogeneus of degree $j_i$. Now we have: $$ f(\lambda x_0,...,\lambda x_n) = \sum_{i=1}^kf_{j_i}(\lambda x_0,...,\lambda x_n)\\ \lambda^d f(x_0,...,x_n) = \sum_{i=1}^k \lambda^{j_i}f_{j_i}(x_0,...,x_n) $$ where the operation in the LHS is the hypotesis and in the RHS we are using the arrow ($\Rightarrow$) of this proposition.

Hence the polynomial $t^d f(x_0,...,x_n) - \sum_{i=1}^k t^{j_i}f_{j_i}(x_0,...,x_n) \in K(x_0,...,x_n)[t]$ has infinite solutions ($K$ is infinite), so it is the $0$ polynomial.

Then in the RHS survives only the degree $d$ part and $f$ is homogeneus.

jacopoburelli
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Menezio
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  • wait how could you possibly use B implies A when we're trying to prove B? Using B implies A suggests to me that B is somehow already true – BCLC Nov 23 '20 at 13:00
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Sorry, could not resist to write this one from geometry, due to its simplicity.

Suppose $ABCD$ is convex quadrilateral with sides $a,b,c,d$. Then $ABCD$ is tangent iff $a+c = b+d$ ($a,c$ are opposite sides).

Proof: Suppose $ABCD$ is tangent, then ... $a+c=b+d$

Now the converse. Say lines $AD$ and $BC$ meet at $E$ and draw an incircle in triangle $ABE$. Say tangent at $C$ meet $AE$ at $D'$. We have to prove $D'=D$. . enter image description here Suppose it is not. But then we have, by already proven part $a+c' = b+(d-x)$ By hypothetis we have also $a+c=b+d$ so $c=c'+x$. But this is impossible because of triangle inequality in $CDD'$. A contradiction.

nonuser
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1

Theorem: Forest $G$ with $n$ nodes is connected iff it has $n-1$ edges.

Proof: Suppose we already prove if direction. Now only if.

Suppose it is not connected. Let $C_1,C_2,...C_k$ be it components and $k\geq 2$. Then, by already proven part, we have $c_i-1$ edges where $c_i= |C_i|$ in each component $C_i$. Now we have $$(c_1-1)+(c_2-1)+...+(c_k-1) = n-1$$ But then we have $$n-k = c_1+c_2+...+c_k-k=n-1$$ A contradiction.

nonuser
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