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.

fractions

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

Last night, I received a Facebook message from a Grade 8 student asking why non-terminating repeating decimals are rational. I am posting the answer here for reference.

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.

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