Blogs, Proofs

Double the Size of that Altar

A very long time ago, around 400 BCE, the Greeks were in an age of intellectual discovery and advancement. Philosophers and mathematicians were finding new and exciting uses of logical and rational arguments. However, this was still a time of uncertainty as war, disease and famine hit people all too regularly. During a particularly disastrous plague, a huge portion of the Greek population was killed. In this time of strife, the people of Delos sought the advice of the Oracle of Delphi on how the plague could be ended. In response, she told them that if they were to double the size of the altar to Apollo, Apollo would intervene causing an end to their problems. The people quickly and dutifully built a new altar. They had indeed double the side lengths of the altar as they believe the had been instructed. However, there problems did not end. They went back to the oracle demanding an explanation. At this she told them that they hadn’t double the size of the altar, they had double the side lengths, hence multiplying the size by 8. The would have to go back and create an altar of the appropriate size before they would receive the promised help from Apollo (see A History of Mathematics for reference).

In the past few posts, All those primes and That’s irrational we have been looking at proofs of questions that are complicated and difficult to understand.  However, we were able to find proofs of these that were simple and elegant.  In the problem of doubling a cube, one of the three famous problems of antiquity, we instead have a question that is easily stated, but the proof becomes extremely abstract and difficult to find.  This is one of those questions that embodies the ideas I talked about in What is most interesting: questions, answer or explanation?  The questions posed here is extremely easy to state, and despite the initial confusion of the people of Delos, is extremely easy to understand: we need to create a cube with twice the volume of an original cube.  Note that since this was to be an altar for Apollo, the volume needed to be exact.  Therefore, the only way the Greeks had to work with exact areas was through the use of a straight-edge and compass.  Hence, we need to be able to double the volume of a cube using only straight-edge and compass.  Furthermore, the answer as to how this can be done is also quite simple to understand: it is impossible.  In trying to prove this answer, or more hopefully disprove it, mathematician toiled for approximately 2200 years, until a proof was finally put forth in the 19th century.

Setting up

In order to keep this to one post, I will forego talking about the failed attempts and much to the work that went into trying to solve this question.  Instead I will focus on the key facts that will be needed to show the result.  I mention this here to point out that it wasn’t just the successful proof that this was impossible that was so important, but rather all of the new topics that were discovered and studied in an attempt to answer the question are also important.

The first thing I will mention is that this is inherently a geometric question.  However, we will be using almost no geometry in providing a proof.  The wonderful thing that Descartes was able to do in mathematics is to present the idea of topics being equivalent, even when there appear to be no similarities.  In particular, he realized that if he was presented with a geometric question, he could convert this geometric question into an algebraic question.  Once you then have an algebraic question, you can use algebraic techniques to find a solution to the problem.  Once you’ve found the solution, you can then convert your solution back into a geometric solution.  Graphically we would have a commutative diagram

\[ \begin{array}{c c c}  \text{Geometry} & \rightarrow & \text{Geometry} \\ \downarrow & & \uparrow \\ \text{Algebra} & \rightarrow & \text{Algebra}. \end{array} \]

In this way, we will convert the original problem into an algebraic problem, find a solution and then convert the solution back to geometric.  The utility of this process has played itself out in numerous ways in mathematics history.  Finding solutions to problems can often be quite difficult in the original statement, so converting the problem to something that is easier is extremely helpful.  For example, this is the process used for word problems throughout your academic career.

If we now look at the problem above, we would have the case that we are given a cube with volume, we’ll say 1.  We then have to find a cube with volume 2.  Therefore, we have to find the corresponding side length.  This would then give use the equations \(x^{3}=2\) where \(x\) is side length.  Hence, the problem reduces to, can we find the \(\sqrt[3]{2}\) using only straight-edge and compass if we are given the length 1.

Constructible Numbers

Constructible numbers are precisely the numbers that can be found using only straight-edge and compass if you are given a unit length, 1.  We have talked about straight-edge and compass constructions in Comparing Fractions with Pythagoras.  We noted there that we were able to do a number of things like drawing circles, lines, parallel lines, perpendicular lines, and other constructions.  In particular, we noted that if we have two given lengths, we can add, subtract, and multiply these lengths.  If the length is non-zero, then we can also divide.  Furthermore if the length is non-negative we can take the square root of the length.  Combining these together, we would say the set of all constructible numbers is closed for the taking of sums, difference, products, quotients (bottom non-zero) and square roots (of non-negative numbers).

On the other hand, if we are to find new distances, we can only do that by finding points through the intersections of lines and circles that we can draw.  That is, given constructible points, we can find a constructible line, and given constructible points and distance, we can find a constructible line.  While there is some arithmetic, if we note that the equation of any circle is given by \((x-h)^{2}+(y-k)^{2}=r^{2}\) and the equation of a line is \(y=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})+y_{1}\) we get that any intersection of these equations can be found using only sums, differences, products, quotients (with bottom non-zero) and square roots (of non-negative numbers).

If we combine these two paragraphs together, we get that we can find all constructible numbers, starting with the number 1, and using sums, differences, products quotients and square roots.

Proof

Here we define a sequence of sets.  We let \(F_{0}=\mathbb{Q}\), that is the rational numbers.  Then \(F_{i}\) is the set consisting of numbers that can be written as \(a+b \sqrt{c}\).  We note that each \(F_{i}\) is closed under all of the operations we outlined, except the taking of square roots.  Then if we take up to one more square root, we end up in \(F_{i+1}\).  Since we get the constructible numbers by closing under square roots, we get that the set of constructible numbers, \(F\), is the union (or limit) of this sets, that is \(F=\bigcup\limits_{i \in \mathbb{N}} F_{i}\).

Now we fall into a proof by contradiction as we have for the past two proof posts and assume that \(\sqrt[3]{2}\) is constructible.

  • Then \(\sqrt[3]{2}\) is in \(F_{n}\) for some \(n\).
  • Therefore, for some \(a,b,c \in F_{n-1}\), we have \( \sqrt[3]{2}=a+b\sqrt{c}\).
  • Hence \[ \begin{align*}2=&(a+b\sqrt{c})^{3} \\ 2=&a^{3}+3a^{2}b\sqrt{c}+3ab^{3}c+b^{3}c\sqrt{c} \end{align*}\].
  • We now have \(2=(a^{3}+3ab^{2}c)+(3a^{2}b+b^{3})\sqrt{c})\).
  • Let \(A=(a^{3}+3ab^{2}c)\) and \(B=(3a^{2}b+b^{3})\).   Then the above is equivalent to \(2=A+B\sqrt{c}\) where \(A,B \in F_{n-1}\).
  • Note that if \(B\) is not \(0\), then \(\sqrt{c}=\cfrac{-A}{B}\) which implies \(\sqrt[3]{2} \in F_{n-1}\).
  • Otherwise \(B=0\).
  • Now, we find that \((a-b\sqrt{c})^{3}=A-B\sqrt{c}\).  Since \(B=0\), we have that \(a-b\sqrt{c}\) is also the cube root of 2.  However, there is only one real cube root of 2, so \(a+b\sqrt{c}=a-b\sqrt{c}\).  Therefore, \(b=0\), which implies that \(\sqrt[3]{2} \in F_{n-1}\).
  • We can continue this indefinitely until we get that \(\sqrt[3]{2}\) must be in \(\mathbb{Q}\).
  • However, if we follow the proof that \(\sqrt{2}\) is not rational from That’s irrational, we can show \(\sqrt[3]{2}\) is not rational.
  • Hence, we have a contradiction, so \(\sqrt[3]{2}\) is not constructible.

Conclusion

Here I would like to take a moment to reflect on what we have done in this proof.  We were presented with a question that we really wanted to solve (because we wanted to end the plague).  Attempt after attempt lead to failure after failure.  Therefore, we either changed the rules and created curves other than lines and circles, or we kept going.  Eventually, we had put the problem to the back of minds and moved on to other things.  Then, after exploring fields and fields extensions, we realized that we could use this seemingly unrelated technique to answer a question that had been plaguing us for over two millennia.  We put the proof together in a way that the work no longer resembled what was original asked, and yet we were able to show that the original problem was indeed impossible to do.

Each of the three problems of antiquity have lead to major discoveries in and connections between areas of studies of mathematics.  While we have had many problems that have challenged mathematics for a very long time, no other problems have remained unanswered longer, or provided more room for exploration than these original three.

I hope you enjoyed the proof and the story that went along with it.  If you did, like the post and share it on social media to let your friends know about it.  Furthermore, I hope you’ll take the time to check out the Shop to see the instructional resources Paula and I have written up.

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