Unsolvability of Quintic Equations

General formulas for roots of quadratic, cubic, and quartic equations can be written in closed form using the following algebraic operations: addition, subtraction, multiplication, division, raising to an integer power, and taking an integer root.

However, roots of quintic equations, \[ax^5 + bx^4 + cx^3 + dx^2 + e x + f = 0,\] cannot be written in closed form using these operations.

My PhD advisor, Ron Larson, had told me that this was one of the questions he was asked on his oral PhD qualifying exam. I knew the fact, but never understood the proof, since it involved math that I was not familiar with.

Fred Akalin presents a nice proof using plenty of interactive demos, visualizations, and not much advanced math.

If $1 = 1 sec ...

If $1 were equal to 1 second, the median US household income per year of $50,000 would correspond to half a day.

This helps puts millions, billions, and trillions into perspective.

Roughly,

• $1 million = 2 weeks
• $1 billion = 32 years
• $1 trillion = 300 centuries (before recorded history)

A trillion is a really large number! 

Sunday Afternoon Fun

I spent a glorious afternoon playing the "can you sketch this picture without lifting your pencil?" game with my daughter. We started with a few standard ones that you can find with a simple Google Image search like:

click on image to enlarge

Soon enough, we just started making up random pictures for each other.

After a while, I told her the first part of the secret (Euler Paths), which tells you whether a solution exists.

• identify all the points (nodes)
• write down the number of lines that meet at each node
• scratch off nodes with an even number number of lines
• count the number of (remaining) "odd" nodes
• if this number is 0 or 2, then it is possible to solve the problem

As an example the following figure,

has 3 even nodes (2, 4, and 2), and 2 odd nodes. Thus, it is possible to sketch this "graph" without lifting your pencil.

Showing existence is only the first part, but the second part is an algorithm to actually solve the problem, once you know it is solvable.

Wikipedia lists two methods (both of which are over a century old): Fleury's and Hierholzer's algorithms. But for "kiddie problems" a simpler rule of thumb works in most cases:

So this is the second secret: start from an odd node! For most cases, there are multiple solutions, and it is hard to go wrong with this simple "suggestion"!

PS: If the number of odd nodes is 0, it doesn't matter where you start, and your start and finish points are exactly the same!