## Proof by Contradiction: Knights and Liars

I am currently reading a local book about mathematical problem solving and I came across with a cute proof demonstrating proof by contradiction.  Actually, I heard a very similar problem ages ago, but I never really thought that it was mathematical and never really bothered to solve it.

Anyway, according to the authors, the problem was based on the lectures given by Oleg Goldberg (an International Math Olympiad gold medalist) at the IDEA MATH program, a weekend extracurricular training on problem solving [in Massachussettes?].  The problem and proof are as follows.

Problem

All natives of the Idea island are either knights or liars. Knights always tell the truth, and liars always lie.  A person says “I am a liar.” Prove that he is not an Idea native.

Note: The idea of putting the image below is to let you do the proof first before scrolling down.  So, no peeking please. » Read more

## Proof Tutorial 2: Proving Square Root of 2 is Irrational by Contradiction

One of the most difficult proof strategies in mathematics is proof by contradiction. If P, for example, is a statement or a conjecture, one strategy to prove that P is true is to assume that P is not true  and find a contradiction so that the statement not P does not hold. If not P does not hold, it follows that P is true.

One well-known proof that uses proof by contradiction is proof of the irrationality of $\sqrt{2}$.  If we consider P to be the statement “$\sqrt{2}$ is irrational”, then not P is the opposite statement or “$\sqrt{2}$ is rational”.  To use proof by contradiction, we assume that $\sqrt{2}$ is rational, and find a contradiction somewhere. If this happens, then we would have shown that $\sqrt{2}$ is indeed irrational.

Before proceeding, recall that a rational number is a fraction with non-zero denominator.  We know that all fractions can be expressed in lowest term.  A fraction in $\displaystyle\frac{a}{b}$ is said to be in lowest term if $a$ and $b$ have no common divisors except $1$.

On the other hand, irrational numbers cannot be expressed as fractions. They are decimal numbers that do not end and do not repeat. For example, $0.10100100010000...$ is an irrational number (the three dots means and so on which means that the number does not end). The most popular irrational number is $\pi$.

Now, we prove our conjecture.

Conjecture: The $\sqrt{2}$ is irrational.

Proof:

Suppose $\sqrt{2}$ is rational, then it can be expressed in fraction form $\displaystyle\frac{a}{b}$ . Let us assume that our fraction is in lowest term, i.e., their only common divisor is $1$. Then,

$\sqrt{2} = \displaystyle\frac{a}{b}$

Squaring both sides, we have

$2= \displaystyle\frac{a^2}{b^2}$

Multiplying both sides by $b^2$ yields

$2b^2= a^2$*

Since $a^2 = 2b^2$, we can conclude that $a^2$ is even because whatever the value of $b^2$ has to be multiplied by $2$. If $a^2$ is even, then $a$ is also even. Since $a$ is even, no matter what the value of $a$ is, we can always find an integer that if we divide $a$ by $2$, it is equal to that integer. If we let that integer be $k$, then $\displaystyle\frac{a}{2} = k$ which means that $a = 2k$.

Substituting the value of $2k$ to $a$ in *, we have $2b^2= (2k)^2$ which means that $2b^2=4k^2$.  Dividing both sides by $2$, we have $b^2 = 2k^2$. That means that the value $b^2$ is even, since whatever the value of $k$ you have to multiply it by $2$.  Again, if $b^2$ is even, then $b$ is even.

This implies that both $a$ and $b$ are even, which means that both the numerator and the denominator of our fraction are divisible by $2$. This contradicts our assumption that $\displaystyle\frac{a}{b}$ has no common divisor except $1$. Since we found a contradiction, our assumption is, therefore, false. Hence, the theorem is true.

Notice that I have highlighted the word suppose and assume in the proof. This is one unique feature of proof by contradiction. You can always assume, most of the time, the opposite of the conjecture as long as the following statements are logically valid.