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Sheep in Wolf's Clothing

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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