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.

Why is any number raised to 0 equals 1?

Why is any number raised to 0 equals 1?

If we raise a number to an exponent, we are multiplying it by itself a certain number of times.  For example, $3^4$ means you have to multiply $3$ by itself $4$ times.   In exponential notation, we call $3$ the base and $4$ the exponent.

Shown below are examples of exponential expressions and their expansion.

$2^3 = 2 \cdot 2 \cdot 2$

$4^7 = 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\cdot 4\cdot4$

$x^4 = x \cdot x \cdot x \cdot x$

$(a + b)^3 = (a+ b)(a+b)(a+b)$

$x^m = x \cdot x \cdot x \cdot \ldots \cdot x \cdot x$ (Multiply $x$ by itself, $m$ times)

The$\ldots$ symbol means “and so on.” It represents $x$’s that are missing.  It is convenient to use the said symbol for large values of $m$.

Multiplying Expressions with Exponents

If we want to multiply expressions with the same base, let us see what happens. For example, what will happen if we multiply $2^3$ and $2^5$?

From above, $2^3 = 2 \cdot 2 \cdot 2$ and $2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$. Multiplying the two expressions, we have

$2^3 \cdot 2^5 = (2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2) = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^8$

From our computation, we can conclude that if we multiply to expressions with the same base, we have just have to add their exponents (Can you see why?). That is, for expressions $x^m$ and $x^n$,

$x^m \cdot x^n = x^{m+n}$ (*)

Q1: What if the base of the two expressions are not the same? Will our formula above still apply?

Dividing Expressions with Exponents

What about dividing expressions with exponents? Suppose, we want to divide $2^5$  by $2^3$.

We know that $2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ and $2^3 = 2 \cdot 2 \cdot 2$. Dividing the two expressions, we have

$\displaystyle \frac{2^5}{2^3}= \displaystyle\frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2} = 2^2$

Since, three $2$‘s are canceled out, we can therefore conclude that in dividing two expressions with the same base, we just have to subtract their exponents. That is, for expressions $x^m$ and $x^n$,

$\displaystyle\frac{x^m}{x^n} = x^{m-n}$. (**)

What happens if the exponent of the denominator is larger? For example, $\displaystyle \frac{2^2}{2^7}$?

From (**), $\displaystyle\frac{2^2}{2^7} = 2^{2-7} = 2^{-5}$.

Now, let us compare this result when we expand our expression:

$\displaystyle \frac{2^2}{2^7}= \displaystyle\frac{2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \frac{1}{2^5}$

Our observation tells us that, $2^{-5} = \displaystyle\frac{1}{2^5}$. Therefore, $x^{-m}= \displaystyle\frac{1}{x^m}$.

Q2: In general, what is the value of $\frac{x^m}{x^n}$ if $m < n$?

What if the exponents of the numerator are equal? For instance, $\frac{2^5}{2^5}$. This is practically dividing the same number, so obviously the answer is $1$. However, we can also use our conclusion above.

From (**), $\displaystyle\frac{2^5}{2^5} = 2^{5-5} = 2^0 = 1$

Conclusion

Here we observe that raising$x$ (or any expression) to $0$  means that the number of factors in the numerator and the number of factors in the denominator is the same. Therefore, $x^0$ can be expressed as $x^{m-m}$ for any value of $m$. But from (**), $x^{m-m}$ is equivalent to $\displaystyle\frac{x^m}{x^m} = 1$

Therefore, any number raised to $0$ (with the exception of 0) equals $1$.

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