Sheep in wolf's clothing
Problem
In the following exhibits we give an advanced or alternative way of thinking about mathematics concepts which are likely to be known in a more familiar form.
Explore these structures and experiment by substituting particular values such as $0, \pm 1$. Can you work out what they represent?
Exhibit A
All pairs of integers such that:
$$(a, b) + (c, d) = (ad+bc, bd)\quad\quad (Na, Nb) \equiv (a, b) \mbox{ for all } N\neq 0$$
Can you find two pairs which add up to give $(0, N)$ or $(0, M)$ for various values of $N$, $M$?
A set of ordered pairs of real numbers which can be added and multiplied such that
$(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 +y_2)$
$(x_1, y_1)\times (x_2, y_2) = (x_1x_2 -y_1y_2, x_1y_2+y_1x_2)$
Exhibit C
A set defined recursively such that
$+_k(1) = +_1(k)$
$+_k(+_1(n)) = +_1(+_k(n))$
$\times_k(1) = k$
$\times_k(+_1(n)) = +_k(\times_k(n))$
In these rules, $k$ and $n$ are allowed to be any natural numbers
Once you have figured out what these structures represent ask yourself this: Are these good representations? What benefits can you see to such a representation? How might familiar properties from the structures be represented in these ways?
Student Solutions
Exhibit A
The condition $$(Na, Nb) \equiv (a, b) \mbox{ for all } N\neq 0$$ ought to give it away: $$(a,b) \iff \frac{a}{b}$$ This statement simply says that if the numerator and denominator of a fraction share a common factor, they can be cancelled down.
Exhibit B
If we represent a complex number a+bi by the ordered pair (a,b), we get the required properties:
$$(a+bi) + (c+di) = (a+c) + (b+d)i \iff (a,b) + (c,d) = (a+c, b+d)$$
$$(a+bi) \times (c+di) = (ac-bd) + (ad+bc)i \iff (a,b) \times (c,d) = (ac-bd, ad+bc)$$
Exhibit C
These formally define addition and multiplication over the natural numbers. Can you see how the familiar properties we're used to follow from them?
The first implies $k+1 = 1+k$, i.e. addition is commutative.
The second implies $k+(1+n) = 1+(k+n)$, i.e. addition is associative.
The third implies $k\times 1 = k$, i.e. 1 is the multiplicative identity.
The fourth implies $k\times(1+n) = k+(k\times n)$, that mulitplication is distributative over addition.
This is a rigorous treatment of a very familiar concept. For more information on this subject, you could start by reading this Wikipedia article.