This week, I have started revisiting past tutorial questions as a preparation for the final exam coming up in less than a month. I noticed that I have had a few issues with problem-solving in the past, especially with problems that were introduced in the tutorial problem sets. As a result, I was reminded of
Polya's problem solving approach which aims to break down a problem into distinct steps as to facilitate the process of finding a solution.
I have chosen to use the principles suggested by Polya to solve the following tutorial problem which I initially found challenging.
Understand the problem
The problem above seems to be about a special relationship between a set of integers: for all integers x and y, should x be less than or equal to y, then there must exist some integer z which is both greater than or equal to x, and less than or equal to y.
In other words, the integer z falls exactly between the integers x and y if the antecedent is valid. The following number-line is a representation to facilitate the visualization of the problem.
Let's say that our x in this case is 1, and y is 7, then according to the above statement, z can be any of the integers between and including 1 and 7. Of course, this is just an example as the above should work for all x and y.
Devise a plan
The consequent of the statement gives us a clear hint. Although z is in between the two integers x and y, it can also be one of the two integers. That is to say, z can be x or y, or something in between. What if we choose z to be equal to x? x that is less than y would remain less than y, and at the same time x will be less than or equal z (in fact, x would equal z), and hence by transitive nature of the greater-than-or-equal sign, x ≤ z ≤ y.
Carry out plan
At this stage, I intend to directly prove the above claim using the plan devised earlier.
Assume x ∈ Z # x is a generic integer
Assume y ∈ Z # y is a generic integer
Assume x ≤ y # assumed antecedent
Let z = x , then z ∈ Z # x is an integer, then so is z
then z ≤ y # since x ≤ y
then x ≤ z ≤ y # transitive nature of the ≤ sign
Then x ≤ y ⇒ ∃x∈ Z, x ≤ z ≤ y # assumed antecedent, got consequent
Then ∀y ∈ Z, x ≤ y ⇒ ∃x∈ Z, x ≤ z ≤ y
Then ∀x ∈ Z, ∀y ∈ Z, x ≤ y ⇒ ∃x∈ Z, x ≤ z ≤ y
Look back
Direct proof the above statement worked without issues! It is quite simple to see the connections once the Polya's principles are followed, mathematical rules are investigated thoroughly and different modes of the problem are investigated. For example, had I not made the visual representation, I quite possibly would not have been able to properly visualize the problem. This is related to another discipline I immensely enjoy which is cognitive science; the brain needs different cues to encode a problem and providing the visual aid along with perhaps auditory explanation of the problem can drastically facilitate the process.
I must admit that it takes me considerably longer to arrive at a solution when I do not separate my work into segments as Polya suggested. This means that organization is the key to a successful problem-solving strategy