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Intuition can only take us so far: Fun with Factors (Part 2)

Govind Sharma
March 9, 2021
February 27, 2024
Table of Contents

Continuing with our series on “Fun with Factors” (please find the first part here), we had another session on “Intuition can only take us so far”, wherein we discussed how non-intuitive concepts such as irrational numbers are very much real. Furthermore, we established the importance of grounding ideas to their bare-bones structure, lest we confuse ourselves and fall into paradoxes.

The Irrational Route

For a number to be rational is to possess the ability of being expressed in the form of a fraction -- or the well-known p-by-q (p/q). Now, just for completeness, recall that ‘p’ and ‘q’ should be integers. And ‘q’ should be non-zero.

That said, is it not easy to see that every number is rational? What’s the big deal? Wait, prepare to be challenged! You need to prove (or disprove) that the square root of 2 (i.e., √2) is a rational number. Oh, I heard you! You say √2 an "imaginary" concept with no practical existence. Smart; you took the challenge to another level! So let’s first see how √2 looks like, and how it’s very real!

Take a square piece of cloth ABCD, each side of which measures 1 m. Now cut it into two pieces along one of its diagonals (say, AC). What you get are two right-angled triangles ABC and A’DC’. Let’s take one of them -- ABC. How much do its sides measure? We know AB = 1 m and BC = 1 m; but AC = ?.

The Irrational Route

Following Pythagoras’ advice, we could compute AC = √(AB² + BC²) = √(1+1) = √2. Bingo! We have a triangular cloth with one side measuring √2 metres. But you might object! “Why √2? I used a ruler and measured it to be 1.414 m.” Are we in a fix? Not yet. Analytically, we have AC = √2, but on measuring it using a ruler, we get 1.414. One can deduce that the value of √2 is 1.414. That is a smart move because if you could prove that, you would have √2 = 1.414 = 1414/1000, a rational number indeed! Let us see.

So what sorcery is this entity called √2? Simply speaking, it’s the number whose square should be 2. So, we should expect the square of 1.414 to be 2. Alas! It turns out that 1.414² = 1.999396, a little short of 2, isn't it?

Never mind, you procure a better ruler with more precise scale markings and measure the diagonal side of the cloth (AC) to be 1.41421356237 m. But on squaring it, we get 1.41421356237² = 1.9999999999912458800169, again, short of 2.

The fact of the matter is that no matter how precisely you measure the value of √2, it’s inexpressible as a fraction. But how do I convince you of that? You should demand a proof. A proof that √2 is not a rational number.

Let’s see what we could do:

Assume √2 to be a rational number; and let’s give this assumption a name: "The Rational Root Assumption" (TRRA). Now, if TRRA were to be true, we should be able to find two integers p and q such that √2 = p / q. In addition, let us demand p and q to meet a condition: that they have no common factors except 1. Let us call this the “no common factors” condition (NCFC). Now, “√2 = p/q” simply means that p = q√2, or p² = 2q². As soon as you multiply something by 2, the product becomes an even number. So we have 2q² to be an even number, and hence p² is an even number as well. This leads to our first conclusion: that p is an even number (because if it were not, then it would be odd, and if it were odd, then p would be equal to 2k+1 for some integer k, and this would mean (2k+1)² = 4k²+4k+1 = 2(2k²+2k) + 1 would be odd, and so would p² be, which is not possible since we showed p² is even). Let’s call it the “p is an even number” conclusion (PENC). But what does PENC mean? That p could be written as 2m for some suitable integer m. Let’s replace this in the equation p² = 2q². We get (2m)² = 2q², or 4m² = 2q² or q² = 2m². Oh, we have seen this before. This means q² is even, and hence q is even (for reasons made clear above). Let us call this the “q is an even number” conclusion (QENC).

The summary of the foregoing discussion is this: [TRRA and NCFC] implies [PENC and QENC]. In other words, if √2 is a rational number with numerator p and denominator q, and p & q have no common factors, then both p and q are even numbers. Wow, isn't that hard to believe, because how could p and q be even and not have any common factors? If they are even, they would have 2 as a common factor. Now, this is what we call a contradiction! And since the logical flow was flawless, there is only one explanation to the contradiction: the TRRA assumption -- that √2 is rational. Hence, we have proved that √2 is irrational. Period!

Was this discussion easy to follow? Yes.

Was it easy to write? No, because we had used wholesome English words to express the proof.

In fact, proofs are best expressed using shorthand symbols. To illustrate, the following would be a shorter version of the same argument:

To prove √2 ∉ .

Proof: Assume √2 ∈ .

⇒ ∃ p, q ∈ with p⊥q and q ≠ 0 s.t. √2 = p/q.

⇒ p² = 2q² ⇒ p²|2 ⇒ p|2 ------------------> (1)

⇒ m ∈ Z s.t. p = 2m ⇒ (2m)² = 2q² ⇒ q² = 2m² ⇒ q²|2 ⇒ q|2 ------> (2)

Now from (1) and (2) above, we have p|2 and q|2.

⇒ p⊥q is not true. Hence, we have a contradiction.

So, √2 ∉ . Hence, proved.

So √2, after all, is an irrational number and hence could not be written as a fraction of two integers.

Impossible Probabilities

To find the probability of an event is to measure something. And the prerequisite to make measurement possible is to define what to measure. Imagine what happens if what you want to measure is not well-defined. When asked to compute the conversion ratio of a campaign, your first question is to seek what the definition of a conversion event is. Let us understand the importance of defining concepts explicitly and clearly with the following example from the book on Probability and Statistics by Vijay K. Rohatgi et al, referred to as one of Bertrand’s paradoxes.

Question: A chord is drawn at random in the unit circle. What is the probability that the chord is longer than the side of the equilateral triangle inscribed in the circle? 

To understand the question more clearly, consider the circle as follows.

A chord is drawn at random in the unit circle

We have a circle (in red) centered at C with radius r = 1. Inscribe into it an equilateral triangle PQR (blue). If we now randomly draw a chord on this circle (call it chord AB), what is the probability that it is longer than the side (say s = PQ = QR = RS) of the triangle PQR?

Do you see any problem in the question formulation? If no, then you might be surprised to know that there are at least three solutions depending on how one defines the concept “a chord at random”.

Solution 1: Every chord on the circle could be uniquely defined by its end-points. Let us fix one of the end-points -- A -- on the circumference of the circle. This also defines a unique inscribed equilateral triangle APQ. The choice of the other end-point (B) dictates the length of the chord AB.

If B lies on the arc between A and P (Case 1 below), we get a chord shorter than the side of the triangle. Similar is the case when B is chosen on the circumference of the circle between A and Q (Case 2 below). But when we choose B to be somewhere on arc PQ (Case 3), we get a longer chord. 

Solution  for  A chord is drawn at random in the unit circle

Hence, we have the favourable points that could act as B (i.e., in a way that AB is longer) to be points on the circumference between points P and Q (Case 3). Now, since points A, P, and Q divide the circumference of the circle into three equal arcs AP, PQ, and AQ. We have length(arc AP) = length(arc PQ) = length(arc AQ) = 2𝜋/3. Hence, we get the desired probability as length(arc PQ) / circumference = (2𝜋/3) / 2𝜋 = 1/3.

Solution 2: Another way in which the length of a random chord is uniquely determined is by the distance of the chord’s midpoint from the circle’s centre. If we fix a radius OC, we would have an equilateral triangle PQR cutting OC at S. Moreover, length(OS) = length(SC) = length(OC) / 2 = 0.5. Our problem could be solved by picking a point X on OC and drawing a perpendicular line AXB as a chord.

Solution2  for  A chord is drawn at random in the unit circle

Now, where that X is picked decides how long the chord would be. If X is picked on line SC, we have a shorter chord; and the same done on line OS gives a longer one. So our favourable region to pick X is line OS. In other words, the desired probability would be length(OS) / length(OC) = 0.5 / 1 = 1/2.

In conclusion, we have that the same question has two solutions -- 1/3 and 1/2 -- based on our interpretation of the concept of a “random chord”. If you refer to the book, there is another solution that gives a probability of 1/4. This shows how important the exercise of “defining” a concept could be.

At Factors, we support the philosophy of crunching numbers (rather than intuition) to provide intelligent marketing insights, which are only a click away for you to experience: click here to schedule a demo with us. To read more such articles, visit our blog, follow us on LinkedIn, or read more about us.

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