Universal and Existential Quantification

    In Week 3, the class covered notes on quantifiers. Fortunately, all of the logic was already taught during the first 2 weeks of MAT137. Hence, following along with the class was easy breezy. 

    The first topic being talked about was universal quantification, represented by the symbol ∀. This ‘upside down A ‘ symbol is defined as “for all” or “for every”. When being used in a statement, ∀ can either make all of the examples in the statement true or make all of the examples false. For example, the statement “all even numbers are real numbers” can be written as ∀x∈{even numbers}, x∈R, which is true. Contrarily, the statement “every real number is even,” (∀x∈R, x∈{even numbers}) is false, since odd numbers and rational numbers also belong to the set of real numbers.

    The second topic being discussed is existential quantification, represented by the symbol ∃. This “flipped E” means “there exists” or “there is at least one.” It is used in statements where there are some examples that are true or some examples are false. For instance, the statement ‘there exists an integer that is greater than zero’ can be written as “∃x∈Z, x>0”. This statement is true because it is basically saying “there are some positive integers”, meaning that there could also be negative integers (a proof for this would be the number 3). Now, if I change the existential quantifier in this statement to the universal quantification so that “all integers are positive” , then my statement would be false since a counter-example would be that -2 is also an integer but it is smaller than 0. Differently, if I say “there is an integer between the number 1 and 2” (∃x∈Z, 1<x<2), then this statement is false since that no integers are between 1 and 2.

    Lastly, the two quantifiers can be combined into a more complex statement. For instance, “for all integers, there exists at least one other integer that is greater than the said integer” (∀x∈Z, ∃y∈Z, x<y) or in simpler words “a biggest integer does not exist”, which is true. However, if I switched the order of the statement and turn it the other way around into “∃y∈Z, ∀x∈Z, x<y”, then the statement is false, since it is basically saying that “there exists one integer that is bigger than every other integer”. In conclusion, the order in which an existential or universal quantification is used is very important, and if applied in the wrong way would make a true statement false.

My struggle on Week 1 materials

        In week 1, I was presented with a new approach to tackling complicated problems in CSC165. It followed a detailed 4 steps, which consisted of understanding the problem, formulating a plan, executing the plan, and finally, verifying the solution. I literally laughed at the steps because it seemed so simple and common sense. As effortless as it may sound, each steps actually requires students to record in a way that makes sense to the person marking the paper and to the writer him/herself. 

        When I was presented with a sample problem named “Streetcar drama,” and was asked to apply the new problem-solving mechanism, I was completely confused. First of all, the problem seemed like a normal “solve for x problem,” however information required to create a system of equations was insufficient. Second, the question was presented in a conversation style and some info seemed irrelevant. Therefore, when professor Heap asked the class to solve on their own, I sat on my chair for the entire time, struggling to jot something down. And when the class was discussing the answer to this problem, I realized that this is not high school math. Other than applying the simple skill of solving equations, CSC165 also requires students to logically deduce information based on sentences involving no numbers. Because the class only discussed the approach to the problem but not the answer, I went home and spent some time thinking about the problem. Guided by the class discussion and Polya’s approach, I first began by reading and understanding the problem:

You are on a streetcar during rush hour, and overhear the following conversation fragments between persons A and B:

A: haven’t seen you in a long time! How old are your three kids now?

B: The product of their ages is 36.

A: That doesn’t really answer my question…

B: Well, the sum of their ages (in years) is —[at this point a fire engine goes by and     obscures the rest of the answer.]

A: That still doesn’t really tell me how old they are.

B: Well, the eldest plays piano.

A: Okay, I see, so their ages are—[at this point you have to get off, and you miss the     answer]

Then, I listed the givens:

1) from the first part of the conversation, I know that B has 3 kids with ages a, b, c where abc = 36

2) since the product of the kids’ ages was 36, it is correct to assume that 1 ≤ a, b, c ≤ 36

3) from the sentence “the eldest plays piano,” I know that there is a child who is older than the other two by at least a year

      Despite the little amount of information I extracted, I followed Polya’s approach and continued on by stating the required, which was to find the ages of the three kids, represented by variables a, b, and c. My next step is to devise a plan that can find the solution. My first instinct was that this problem is similar to system of equations, where I can solve for variables when given both the product and sum of the numbers. So I carried out my plan by writing down all the possible three factors between 1 and 36 that can multiply to 36.

        After listing all the possible combinations for a, b, and c, I faced another roadblock — I don’t know the sum of the kids’ ages. With only one equation, the three variables cannot be solved. Applying Polya’s approach, I reverted back to step three and determined what info I was missing. And from there, I changed my original plan by rereading the conversation. After many tries I realized that the two younger kids are probably twins. This is because when A was given the product and the sum of the ages, and was told that the eldest plays the piano, he is able to arrive at the conclusion. The only way this could be possible is that the two younger ones are twins. Anything else becomes impossible to solve since it will involve three distinct variables and only two equations can be created. Furthermore, the sum of the ages is not unique, meaning that at least 2 combinations of ages result in the same sum. This is because even after given the sum and the product, A still couldn’t figure out the combination, since there is more than one possibility. And when given the hint that the oldest plays the piano, A can quickly eliminate one of the solution since the two younger ones have to be twins. So I began writing down the corresponding sum to each combination. It turns out that both 1, 6, 6 and 2, 2, 9 result in a sum of 13. However, 1, 6, 6 does not work because there is only one eldest child. Hence 2, 2, 9 is the only solution to the kids’ ages.

        In conclusion, although I struggled to understand Week 1’s lessons, and was frustrated with my intelligence, I was able to overcome the difficulties. Not only did I familiarized myself with the problem solving mechanism, I also gained insight into a new type of math problem that involves logic and reasoning. Even though I still don’t feel confident about the course materials, I believe through some extra help and hard work, I will gradually become accustomed to this new subject.