Thread: 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

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 . This set is defined as the set of subsets of
where each subset contains just two elements, whose sum is . In other words:

(i) If a function is one-to-one, then means that . Now takes an element of and maps it onto the subset of that contains . For instance:

But also:

So ...?

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

Grandad

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

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 , not just a specific one.

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

Grandad