Musings on mathematics and teaching.

Month: August, 2013

Ice cream quiz

This problem was posted in an ice cream shop in Oxford, England. I found it surprisingly difficult.

Here is my solution. Suppose that $x + 3 = a^3$ and $x^2 + 3 = b^3$. After substituting,

$a^6 > (a^3 - 3)^2 + 3 = b^3$

which implies that $a^2 - 1 \ge b$. Therefore

$(a^3 - 3)^2 + 3 \le (a^2 - 1)^3$

$a^6 - 6a^3 + 12 \le a^6 - 3a^4 + 3a^2 - 1$

$3a^4 - 6a^3 - 3a^2 \le -13$

$3a^2 (a^2 - 2a - 1) \le -13$

But the left side is positive for $a \ge 3$, so there are no whole number solutions.

Image credit: Janet McKnight

Thanks to @samuelprime and @nodrogadog for pointing out mistakes in earlier versions of this post.

One thousand

• People: moderately sized school.
• Dollars: price of a laptop computer.
• Pounds: weight of a polar bear.
• Miles: distance from St. Paul to Dallas.

One million

• People: Population of San Jose, California. Participants in the  Million Man March.
• Dollars: Cost of an expensive house or a CAT scanner.
• Pounds: Maximum take-off weight of a 747.
• Miles: Lifetime distance traveled by car (100 miles/day for 30 years).

One billion

• People: Population of India.
• Dollars: Net worth of JK Rowling, author of the Harry Potter books.
• Pounds: The Golden GateBridge weighs about 2 billion pounds.
• Miles: Distance to Saturn.

One trillion

• People: The hypothetical populations of the world, if all land area on the globe were as densely populated as Hong Kong.
• Dollars: Approximate cost of the Iraq war.
• Pounds: World annual rice production.
• Miles: One fifth of a light year.
• Seconds: About 30,000 years. The earliest known cave paintings are about 30,000 years old.

• People: Number of ants in the world.
• Dollars: Total world purchasing power for 15 years, at current levels of production.
• Pounds: Total world coal reserves.
• Miles: About 170 light years.
• Seconds: About 30 million years.

Study guide for beginning algebra

Here is a study guide that I wrote when I was a TA for a beginning algebra class. I hope that someone finds it useful. algebra_review

Primes of the form 6k+1

Euclid’s proof of the infinitude of primes is justly famous. Here is one version of this proof:

Let P be a finite set of primes, and let N be the product of the numbers in P. Then N+1 is not divisible by any number in P, since it leaves a remainder of 1. But N+1 must be divisible by at least one prime, so P cannot contain all of the primes. Therefore the set of primes is infinite.

A variation of this argument shows that there are infinitely many primes of the form $6k - 1$.

Let P be a finite set of primes of the form $6k - 1$, and let N be the product of the primes in P. Consider the number $6N - 1$. It is not divisible by 2 or 3, nor is it divisible by any number in P. But it is not possible that all prime factors of $6N - 1$ have the form $6k + 1$, because the product would have the form $6k + 1$ as well.

Therefore, $6N - 1$ has at least one prime factor of the form $6k - 1$ that does not belong to P. Therefore, P cannot contain every prime of the form $6k - 1$, so the set of primes of this form is infinite.

It takes a little bit more work to prove that there are infinitely many primes of the form $6k + 1$. Here is a proof of that fact.

Let P be a finite set of primes of the form $6k + 1$, and let N be a number that is divisible by every number in P. Assume that N is also divisible by 6. Let p be a prime divisor of $N^2-N+1$.

Note that $(N^2-N+1)(N+1)=N^3+1$. so p divides $N^3+1$, or in other words $N^3 \equiv -1 \pmod{p}$ and so $N^6 \equiv 1 \pmod{p}$.

Recall that the order of N modulo p is the least positive k so that $N^k \equiv 1 \pmod{p}$. The order must divide 6. so k = 1, 2, 3, or 6. But $N^3 \equiv -1 \pmod{p}$, so the order cannot be 1 or 3.

Can the order be 2? If $N^2 \equiv 1 \pmod{p}$ and $N^3 \equiv -1 \pmod{p}$ then $N \equiv -1 \pmod{p}$. This would be bad, because then p would divide both $N+1$ and $N^2-N+1$; but $\gcd(N+1,N^2-N+1) = \gcd(N+1,3) < p$, contradiction.

Thus N has order 6 mod p, and the group of units mod p has order $p-1$, so 6 divides $p-1$, which means that p has the form $6k+1$. Therefore, P does not contain all primes of the form $6k+1$, so the set of primes of this form is infinite.

Cancellation in finite groups

In this note, I will state a proof of the following theorem: If $A,\ B,\ C$ are finite groups, and if $A \times B$ is isomorphic to $A \times C$, then $B$ is isomorphic to $C$.

Let $h(X,Y)$ be the number of homomorphisms from $X$ to $Y$, and let $m(X,Y)$ be the number of monomorphisms from $X$ to $Y$.

If $D$ is a finite group then

$h(D,A\times B)=h(D,A)\cdot h(D,B)$ and
$h(D,A\times C)=h(D,A)\cdot h(D,C)$.

But $h(D,A\times B)=h(D,A\times C)$ since $A\times B$ and $A\times C$ are isomorphic. Therefore $h(D,A)\cdot h(D,B) = h(D,A)\cdot h(D,C)$. Dividing both sides by $h(D,A)$ yields $h(D,B)=h(D,C)$.

I claim that $m(D,B)=m(D,C)$ for every finite group $D$. The proof is by induction on the order of $D$. If $D$ has order 1 then $m(D,B)=1$ and $m(D,C)=1$, so the equality holds.

Suppose that $m(E,B)=m(E,C)$ for every group $E$ of order less than $n$, and let $D$ be a group of order $n$.

Then

$h(D,B) = m(D,B) + \sum_K m(D/K,B)$ and

$h(D,C) = m(D,C) + \sum_K m(D/K,C),$

where $K$ ranges over all nontrivial normal subgroups of $D$. But $h(D,B)=h(D,C)$ by the preceding argument, and $m(D/K,B)=m(D/K,C)$ by the induction hypothesis. Therefore $m(D,B)=m(D,C)$.

Setting $D=B$ and $D=C$ in the above yields $m(B,C)=m(B,B)\geq 1$ and $m(C,B)=m(C,C)\geq 1$. Thus there exist monomorphisms $B\rightarrow C$ and $C\rightarrow B$. Since $B$ and $C$ are finite, this implies that $B$ and $C$ are isomorphic. QED.

I am very fond of this theorem because it was a key lemma in my doctoral dissertation. I learned of this argument from Fred Galvin, but I do not know if it was original to him.

The theorem would be false if the word ‘finite’ were omitted. For example, if A is a free abelian group of infinite rank, B is an infinite cyclic group, and C is the trivial group, then $A\times B$ and $A \times C$ are both isomorphic to A, but B and C are not isomorphic to each other.

How much water is in the oceans?

Nat Banting posed the following problem on Twitter.

We could just look up the answer on Wikipedia, but to estimate the answer, we need to know the following quantities:

f = Fraction of Earth’s surface covered by the oceans
d = Mean depth of the oceans

Recall that the surface area of a sphere of radius r is

$S = 4\pi r^2$

so the total area covered by the oceans is

$A = 4\pi r^2 f$.

To approximate the total volume of the oceans, we multiply the area by the mean depth.

$V \approx 4\pi r^2 fd$

Now, let’s plug in some numbers. The radius of Earth is approximately 6400 km, and about 71% of Earth’s surface is covered by the oceans. The mean depth of the oceans is 3.7 km. So the total volume of the oceans is

$V \approx 4\pi \cdot (6400 \text{ km})^2 \cdot 0.71 \cdot (3.7 \text{ km})$

which is about 1.35 billion cubic kilometers.

Let’s express this in more familiar units. Recall that a liter is the volume of a cube that is 10 cm on each side. A cubic meter is 1000 liters (since 10 x 10 x 10 = 1000), and a cubic kilometer is 1,000,000,000 cubic meters (since 1000 x 1000 x 1000 = 1,000,000,000). So a cubic kilometer is equal to one trillion liters. This implies that the volume of the oceans is about 1.35 billion trillion liters, or about 350 million trillion US gallons.