Hello brumby_3 Originally Posted by

**brumby_3** **Let S = {3,4,5,6,7,8,9,10,11,12}. Someone gives you six numbers a1,....a6 from S.**

Let P = {{x,y}|x,y S and x+y =15}.

i) If f: {a1,....a6} --> P is defined by f(ai) equals the set containing ai, explain why f is not one-one.

ii) If ai does not equal aj, but f(ai)= f(aj), show that ai + aj = 15.

iii) Explain why, if 6 numbers are chosen from s, then the sum of some pair of them is 15

First, let's identify the set $\displaystyle \displaystyle P$. This set is defined as the set of subsets of

$\displaystyle \displaystyle S$ where each subset contains just two elements, whose sum is $\displaystyle \displaystyle 15$. In other words:

$\displaystyle \displaystyle P=\{\{3, 12\},\{4,11\},\{5,10\},\{6,9\},\{7,8\}\}$

(i) If a function is one-to-one, then $\displaystyle \displaystyle f(a_i)=f(a_j)$ means that $\displaystyle \displaystyle a_i = a_j$. Now $\displaystyle \displaystyle f$ takes an element $\displaystyle \displaystyle a_i$ of $\displaystyle \displaystyle S$ and maps it onto the subset of $\displaystyle \displaystyle P$ that contains $\displaystyle \displaystyle a_i$. For instance:

$\displaystyle \displaystyle f(4) = \{4,11\}$

But also:

$\displaystyle \displaystyle f(11) = \{4,11\}$

So ...?

Can you complete (ii) and (iii) now?

Grandad