Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm looking at "Basic Set Theory" by A. Shen. The very first 2 problems are: 1) can the oldest mathematician among chess players and the oldest chess player among mathematicians be 2 different people? and 2) can the best mathematician among chess players and the best chess player among mathematicians be 2 different people? I think the answers are no, and yes, because a person can only have one age, but they can have separate aptitudes for chess playing and for math. Is this correct?

share|improve this question
add comment

2 Answers

up vote 10 down vote accepted

Yes, it’s correct. If $M$ is the set of mathematicians, and $C$ is the set of chess players, you’re looking rankings of the members of $M\cap C$. If for $x\in M\cap C$ we let $m(x)$ be $x$’s ranking among mathematicians, $c(x)$ be $x$’s ranking among chess players, and $a(x)$ be $x$’s age, then there is a unique $x_a\in M\cap C$ such that $$a(x_a)=\max\{a(x):x\in M\cap C\}\;,$$ but there can certainly be distinct $x_m,x_c\in M\cap C$ such that $$m(x_m)=\max\{m(x):x\in M\cap C\}$$ and $$c(x_c)=\max\{c(x):x\in M\cap C\}\;.$$

All of which just says what you said, but a bit more formally.

share|improve this answer
    
Thanks very much for the help you guys! –  Peter Jun 11 '12 at 5:06
    
So what's $V$ in the first displayed equation? –  Asaf Karagila Mar 18 '13 at 22:07
    
@Asaf: Right next to the $C$ on my keyboard. –  Brian M. Scott Mar 18 '13 at 22:08
add comment

(1) Think of it in terms of sets. Let $M$ be the set of mathematicians, $C$ the set of chess players. Both are asking for the oldest person in $C\cap M$.

(2) Absolutely fantastic reasoning, though perhaps less simply set-theoretically described.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.