14.1.6. Question Number 11

In question 11, we are supposed to use strong induction to prove that √2 is irrational. Just to recap we already proved that √2 is irrational using a proof by contradiction using proof by contradiction, but here I am asking you to do the same thing to show the same thing using strong induction so before starting the strong induction proof we have to first identify the universal statement which we are trying to make.

In this chunk, we introduce the goal of question 11, which is to prove that the square root of 2 (√2) is irrational using strong induction. Recall that irrational numbers cannot be expressed as a fraction of two integers. We mention that we have previously proved this using proof by contradiction and now want to attempt it using a different method, namely strong induction. Before we proceed, it’s essential to identify the universal statement regarding √2 that we aim to prove.

Think of proving that a number is irrational like proving there’s no exact recipe for a cake that meets all dietary restrictions. If you’ve already shown it can’t fit the rules precisely (proof by contradiction), now you're trying to illustrate that no matter the number of ingredients (using strong induction), it still won’t work out perfectly every time.

Remember, an induction is used to prove a universally quantified predicate. So first we have to identify what exactly is the predicate here. So the predicate P(n) here is the following; P(n) is the predicate that √2 is not equal to n/b for any positive integer b and I want to prove that this universal quantification is true using strong induction.

Here, we define the specific statement we intend to prove through strong induction. The predicate P(n) states that √2 cannot be expressed as the fraction n/b, where n is any positive integer and b is another positive integer. We then provide clarity that this universal quantification is fundamental in establishing that √2 is irrational, as it shows that there is no integer pair (n, b) that can express √2 as a fraction.

Imagine declaring that no matter what bakery you go to (representing the various integers), none can give you a perfectly rational slice of cake (√2). The predicate is the assurance that every time you try, and every place you go, it remains true.

I start with the base case. I start with the base case, my base case will be when n is equal to 1 that means √2 is not equal to one over any b where b is a positive integer and this is obviously true because we know that the value of √2 is greater than one and one over any positive integer will be strictly less than or equal to one.

In this part, we establish the base case for our strong induction proof. We take n = 1 and demonstrate that √2 is not equal to 1/b for any positive integer b. Since √2 is approximately 1.414, it is evident that this fraction is always less than √2. This confirms our base case, reinforcing the truth of our predicate P(1).

Consider comparing the height of a toddler (representing 1/b) to that of an adult (√2). No matter how tall the toddler grows, they will always be shorter than the adult, illustrating how √2 remains above all fractions of the form one over a positive integer.

Now assume my inductive hypothesis is true, that means √2 cannot be represented in the form of one over b, √2 cannot be represented in the form of 2 over b and in the same way √2 is not cannot be represented in the form of k over b.

In this segment, we state our inductive hypothesis. We assume that for some integer k, √2 cannot be represented as n/b for any integers less than or equal to k. This sets up the structure for our inductive step, where we will prove it for k + 1.

Picture this like climbing stairs where if you know the rule holds true for steps 1 to k (your inductive hypothesis), you are preparing to show that it applies to step k + 1, ensuring the rule remains true for every step upwards.

Now to prove the statement is true for k + 1, I will be using a proof by contradiction and that is allowed because overall I am using an inductive proof mechanism where I have to now prove that this proposition P(k + 1) is also true, that I can prove using contradiction with the help of induction.

The focus here shifts to proving the statement for k + 1. We assume for contradiction that √2 can be expressed as (k + 1)/b for some positive integer b. By substituting this into the known irrationality proof of √2, we can derive further consequences. This indirect method serves to demonstrate that the assumption must be false, thus concluding that the original statement holds.

Think of it like testing a theory that says a particular fish cannot exist in a specific ocean. If you theorize a new fish exists (the contradiction) and then examine its characteristics, you show through exploration that it cannot fit within the known parameters of ocean life.

So I can say that √2 can be represented in the form 2s / 2t, two cancels out and I get the conclusion that √2 is of the form s/t where s is less than equal to k, this is because I started with √2 equal to k + 1 and k + 1 is 2 s.

In the concluding steps, we show that through our contradiction, if √2 were represented in a simplified form (where the 2s and 2t cancel), we arrive at a contradiction against our inductive hypothesis that such a representation does not exist. This confirms that there is no integer fraction possible, thus proving √2 is indeed irrational.

Returning to our fish analogy, once you establish your new fish doesn’t fit the known characteristics of ocean life, it leads to the conclusion that such a fish type does not exist, just like we conclude the illusory representation of √2 as rational does not hold true.

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