# Tag Archives: rational numbers

## Are all fractions rational numbers?

No.

A rational number can be expressed in the form $\displaystyle\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. In other words, it is a fraction whose denominator is not zero, and both the denominator and numerator are integers.

Some fractions, however, may contain a numerator or denominator that is not an integer. Some examples of such fractions are

$\displaystyle\frac{\sqrt{3}}{2}$, $\displaystyle\frac{\pi}{4}$ and $\displaystyle\frac{e}{2}$.

A rational number may be represented in many ways, but it can always be expressed as a fraction. For instance, $10^{-1}$ is a rational number because we can express it as $\frac{1}{10}$. Also, the number $0.333 \cdots$, a repeating decimal, is  a rational number because we can also express it as fraction $\frac{1}{3}$.

## Why are Non-terminating, Repeating Decimals Rational

Rational numbers is closed under addition. That is, if we add two rational numbers, we are guaranteed that the sum is also a rational number. The proof of this is quite easy, so I leave it as an exercise for advanced high school students.

Before discussing non-terminating decimals, let me also note that terminating decimals are rational. I think this is quite obvious because terminating decimals can be converted to fractions (and fractions are rational). For example, $0.842$ can be expressed as

$\displaystyle\frac{842}{1000}$.

Further, terminating decimals can be expressed as sum of fractions. For example, $0.842$ can be expressed as

$\frac{8}{10} + \frac{4}{100} + \frac{2}{1000}$.

Since rational numbers is closed under addition, the sum of any number of fractions is also a fraction. This shows that all terminating decimals are fractions.  Continue reading

## Proof that log 2 is an irrational number

Before doing the proof, let us recall two things: (1) rational numbers are numbers that can be expressed as $\frac{a}{b}$ where $a$ and $b$ are integers, and $b$ not equal to $0$; and (2) for any positive real number $y$, its logarithm to base $10$ is defined to be a number $x$ such that $10^x = y$. In proving the statement, we use proof by contradiction.

Theorem: log 2 is irrational

Proof:

Assuming that log 2 is a rational number. Then it can be expressed as $\frac{a}{b}$ with $a$ and $b$ are positive integers (Why?).  Then, the equation is equivalent to $2 = 10^{\frac{a}{b}}$. Raising both sides of the equation to $b$, we have $2^b = 10^a$. This implies that $2^b = 2^a5^a$.  Notice that this equation cannot hold (by the Fundamental Theorem of Arithmetic) because $2^b$ is an integer that is not divisible by 5 for any $b$, while $2^a5^a$ is divisible by 5. This means that log 2 cannot be expressed as $\frac{a}{b}$ and is therefore irrational which is what we want to show.