# Thread: Set help appreciated!

1. ## Set help appreciated!

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

Sorry if I've typed it out in a confusing way, the € means "is an element of" lol

2. 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 $\displaystyle P$. This set is defined as the set of subsets of
$\displaystyle S$ where each subset contains just two elements, whose sum is $\displaystyle 15$. In other words:
$\displaystyle P=\{\{3, 12\},\{4,11\},\{5,10\},\{6,9\},\{7,8\}\}$

(i) If a function is one-to-one, then $\displaystyle f(a_i)=f(a_j)$ means that $\displaystyle a_i = a_j$. Now $\displaystyle f$ takes an element $\displaystyle a_i$ of $\displaystyle S$ and maps it onto the subset of $\displaystyle P$ that contains $\displaystyle a_i$. For instance:
$\displaystyle f(4) = \{4,11\}$
But also:
$\displaystyle f(11) = \{4,11\}$
So ...?

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

Grandad

3. They're not one-for-one because one if f(4) and the other is f(11) and they're not the same number, right - one is 4 and one is 11.

For ii), you've pretty much wrote it, f(ai) and f(11) are the same, and do equal 15.

For iii) So you choose six numbers, you can see it looking at the set P, there's always going to be one pair that equals 15 because there will always be a match, say if i choose 3, 4, 5, 6, 7, and 12... 3 and 12 match, or if I choose 5, 6, 7, 8, 9, 10, 11.. then 7 + 8 = 15.

Thanks for your help and please tell me if my answers are right or a bit sketchy

4. Hello brumby_3
Originally Posted by brumby_3
They're not one-for-one because one if f(4) and the other is f(11) and they're not the same number, right - one is 4 and one is 11.

For ii), you've pretty much wrote it, f(ai) and f(11) are the same, and do equal 15.

For iii) So you choose six numbers, you can see it looking at the set P, there's always going to be one pair that equals 15 because there will always be a match, say if i choose 3, 4, 5, 6, 7, and 12... 3 and 12 match, or if I choose 5, 6, 7, 8, 9, 10, 11.. then 7 + 8 = 15.

Thanks for your help and please tell me if my answers are right or a bit sketchy
Yes - they are a bit sketchy for (ii) and (iii), although there isn't much more to do for (ii). I think you'd need to mention a general value of $a_i$, not just a specific one.

For (iii) it would be good to mention the fact that you're using the pigeonhole principle.

Grandad