How to solve set problems without the aid of Venn diagrams?


In a school 70% of students like hamburgers, 60% like pizza. 50% of them like both, hamburger and pizza. How many percents like neither hamburger or pizza?

With the aid of a Venn diagram this problem is piece of cake but I'm having trouble creating a system of equations from these data. So can this problem be solved algebraically? How?

The set theory and algebra look like apple and oranges. Am I wrong?

Note: I'm studying mathematics at high-school level but any explanation would help, no matter how abstract/advanced it is.


Question :

How many percents like neither hamburger or pizza?

Answer :


If 50% of them [students] like both, hamburger and pizza, there are 50% left. Of this 50%, 20% likes only hamburgers (70-50) : left 30%. Of this, 10% likes only pizza (60-50) : left 20%.

In "algebraic" form [but it is quite "offensive" to call it algebraic ...], set the total "population" to $100$.

It is composed by those which likes hamburger [70%] + (plus) those which likes pizza [60%] - (minus) those which likes both [50% - in order to avoid counting twice] + (plus) those which do not likes nothing [which is the unknown].

In formula :

$100 = 70 + 60 -50 +x$;

solve for $x$ to get the result.

  • $\begingroup$ As I've said "With the aid of a Venn diagram this problem is piece of cake". So I know hot to use the set theory to solve the problem. I will rephrase the question: can the problem be solved algebraically? $\endgroup$ – Eugen Mihailescu Jun 11 '14 at 8:38
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    $\begingroup$ It seems to me that you've use the set theory to model your algebraic solution. My question was not addressed exclusively for that particular problem (example). I should mention that I have also tried to find such algebraic equations but always I had to use what I know about sets in order to model an algebraic solution. So it's like a vicious circle. It seems rather hard to model algebraically such problems if you (yet) have no knowledge about set theory. $\endgroup$ – Eugen Mihailescu Jun 11 '14 at 8:55
  • $\begingroup$ @EugenMihailescu - it seems to me that it is very poor "algebraic" ... set four variables : h (ham's lovers), p (pizza's lovers), b (for both) and x (the unknown). We have $h=70$, $p=60$, $b=50$ and $h+p-b+x=100$. Four variables, four equations. We have complicated it enough ? $\endgroup$ – Mauro ALLEGRANZA Jun 11 '14 at 9:03
  • $\begingroup$ that's convincing, thanks. I'm just wondering if someone could model such a problem without knowing anything about set theory. Because I think that you've chosen this model because involuntary you still think in terms of set theory. And this knowledge helps you to find such a model. But what's the probability that a student in the secondary school find such a model if he/she never thought in terms of exclusion/inclusion/etc. $\endgroup$ – Eugen Mihailescu Jun 11 '14 at 9:13
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    $\begingroup$ @MauroALLEGRANZA my position is that your solution is using the inclusion-exclusion principle when you say: in order to avoid counting twice. Once you assume that the equation given by this principle is true, then the problem becomes algebraic. Notice that this principle is about finite sets and can be understood without knowning anything about general set theory (let alone Cantor). $\endgroup$ – Emanuele Paolini Jun 11 '14 at 10:11

The problem is inherently set theoretic. So you don't need Venn diagrams, but you need to use the properties of sets. Here the property you want to use is about cardinality of sets: $$ |A \cup B| = |A| + |B| - |A\cap B|. $$ This is a simple case of the inclusion-exclusion principle. $|X|$ denotes the number of elements (or cardinality) of the finite set $X$.

I think you cannot solve the problem without using set theory at all. This because the statement of the problem can possibly be translated in a mathematical statement only by using set theory. Any purely algebraic solution could use the above formula (which is algebraic if we give a name to the cardinality of each set) but cannot explain why such a formula is true.

  • $\begingroup$ Oh, I know all of these. But this way of solving the problem involves also the concept of a set. Perhaps I should rephrase the question: can the problem be solved algebraically? Btw: thanks for your help. $\endgroup$ – Eugen Mihailescu Jun 11 '14 at 8:35
  • $\begingroup$ I would say no: because the statement of the problem itself can only be modeled by using sets. A purely algebraic solution cannot explain why the formula used is true. $\endgroup$ – Emanuele Paolini Jun 11 '14 at 8:39
  • $\begingroup$ The first sentence is wrong. This problem is inherently not set theoretic. It is combinatorial. Not every problem with sets is set theoretic. $\endgroup$ – Asaf Karagila Jun 11 '14 at 8:43
  • $\begingroup$ @AsafKaragila I would say that combinatorics is based on set theory... I cannot imagine an axiomatization of combinatorics which does not use sets. Instead I can imagine an axiomatization of numbers (real or natural) without sets. $\endgroup$ – Emanuele Paolini Jun 11 '14 at 8:51
  • $\begingroup$ (1) I don't know any axiomatization of combinatorics, compared to calculus, algebraic objects and so. Combinatorics deal with counting. (2) Surely Peano axioms can prove the majority of theorems we see in combinatorics (even if not all of them). (3) By that argument category theory is inherently set theoretic as well? :-) $\endgroup$ – Asaf Karagila Jun 11 '14 at 8:55

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