Is 0.999… really equal to 1?

Introduction

Yes it is. 0.999…  is equal to 1.

Before we begin our discussion, let me make a remark that the symbol “…” in the decimal 0.999… means that the there are infinitely many 9’s,  or putting it in plain language, the decimal number has no end.

For non-math persons, you will probably disagree with the equality, but there are many elementary proofs that could show it, some of which, I have shown below. A proof is a series of valid, logical and relevant arguments (see Introduction to Mathematical Proofs for details), that shows the truth or falsity of a statement.

Proof 1

\frac{1}{3} = 0.333 \cdots

\frac{2}{3} = 0.666 \cdots

\frac{1}{3} + \frac{2}{3} = 0.333 \cdots + 0.666 \cdots

\frac{3}{3} =0.999 \cdots

But \frac{3}{3} = 1, therefore 1 =0.999 \cdots

Proof 2

\frac{1}{9} = 0.111 \cdots
Multiplying both sides by 9 we have

1 = 0.999 \cdots

Proof 3

Let x = 0.999 \cdots

10x = 9.999 \cdots

10x - x = 9.999 \cdots - 0.9999 \cdots

9x = 9

x = 1

Hence, 0.999 \cdots = 1

Still in doubt?

Many will probably be reluctant in accepting the equality 1 = 0.999 \cdots because the representation is a bit counterintuitive.  The said equality requires the notion of the real number system, a good grasp of the concept of limits, and knowledge on infinitesimals or calculus in general.  If, for instance,you have already taken sequences (in calculus), you may think of the 0.999 \cdots as a sequence of real numbers (0.9, 0.99, 0.999,\cdots). Note that the sequence gets closer and closer to 1, and therefore, its limit is 1.

Infinite Geometric Sequence

My final attempt to convince you that 0.999 \cdots is indeed equal to 1 is by the infinite geometric sequence. For the sake of brevity, in the remaining part of this article, we will simply use the term “infinite sequence” to refer to an infinite geometric sequence.  We will use the concept of the sum of an infinite sequence, which is known as an infinite series, to show that 0.999 \cdots = 1.

One example of an infinite series is \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots.  If you add its  infinite number of terms, the answer is equal to 1. Again, this is counterintuitive.

How can addition of numbers with infinite number of terms have an exact (or a finite) answer?

There is a formula to get the sum of an infinite geometric sequence, but before we discuss the formula, let me give the geometric interpretation of the sum above. The sum \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots can be represented geometrically using a 1 unit by 1 unit square as shown below. If we divide the square into two, then we will have two rectangles, each of which has area \frac{1}{2} square units. Dividing the other half into two, then we have three rectangles with areas \frac{1}{2}, \frac{1}{4}, \frac{1}{4} square units. Dividing the one of the smaller rectangle into two, then we have four rectangles with areas \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}. Again, dividing one of the smallest rectangle into two, we have five rectangles with areas \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, and \frac{1}{16} Since this process can go on forever, the sum of all the areas of all the rectangles will equal to 1, which is the area of the original square.

Now that we have seen that an infinite series can have a finite sum, we will now show that 0.999 \cdots can be expressed as a finite sum by expressing it as an infinite series. The number 0.999 \cdots can be expressed as an infinite series 0.9 + 0.09 + 0.009 + \cdots. Converting it in fractional form, we have  \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots.

We have learned that the sum of the infinite series with first term \displaystyle a_1 and ratio r is described by \displaystyle\frac{a_1}{1-r}. Applying the formula to our series above, we have

\displaystyle\frac{\frac{9}{10}}{1-\frac{1}{10}} = 1

Therefore, the sum our infinite series is 1.

Implication

This implication of the equality 0.999 \cdots =1 means that any rational number that is a non-repeating decimal can be expressed as a repeating decimal. Since 0.999 \cdots =1, it follows that 0.0999 \cdots =0.1, 0.00999 \cdots=0.01 and so on. Hence, any decimal number maybe expressed as number + 0.00…01. For example, the decimal 4.7, can be expressed as 4.6 + 0.1 = 4.6 + 0.0999 \cdots = 4.6999 \cdots. The number 0.874 can also be expressed as 0.873 + 0.001 = 0.873 + 0.000999 \cdots = 0.873999 \cdots

Conclusion

Any of the four proofs above is actually sufficient to show that 0.999 \cdots = 1.  Although this concept is quite hard to accept, we should remember that in mathematics, as long as the steps of operations or reasoning performed are valid and logical, the conclusion will be unquestionably valid.

There are many counterintuitive concepts in mathematics and the equality 0.999 \cdots = 1 is only one of the many.  In my post, Counting the Uncountable: A Glimpse at the Infinite, we have also encountered one:   that the number of integers (negative, 0, positive) is equal to the number of counting numbers (positive integers) and we have shown it by one-to-one pairing. We have also shown that the number of counting numbers is the same as the number of rational numbers. Thus, we have shown that a subset can have the same element as the “supposed” bigger set.  I guess that is what makes mathematics unique; intuitively, some concepts do not make sense, but by valid and logical reasoning, they perfectly do.

Notes:

  1. You can find discussions about 0.999… = 1 here and here.
  2. There is another good post about it here and here.
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Division by Zero

In studying mathematics, you have probably heard that division of zero is undefined. What does this mean?

Since we do not know exactly what is the answer when a number is divided by 0, it is probably reasonable for us to examine the quotient of a number that is divided by a number that is close to 0.

If we look at the number line, the numbers close to 0 are numbers numbers between –1 and 1.

Figure 1 – The number line showing the numbers close to 0.

For instance, several positive numbers close to 0 and less than 1 are 0.1, 0.01, 0.001 and so on. Similarly, negative numbers close to 0 but greater than – 1 are –0.1, -0.01, -0.001 and so on.

The table and the numbers below shows the quotient 1/x when 1 is divided by x, where the x’s are numbers close to 0.

Figure 2 – The value of 1/x as x approaches 0 from both sides.

In the graph, as x approaches 0 from the right (as x, where x are positive numbers, approach 0), the quotients of 1/x are getting larger and larger. On the other hand, as x approaches 0 from the left (as x, where x are negative numbers, approach 0), 1/x is getting smaller and smaller. Hence, there is no single number that 1/x approach as x approaches 0.  For this reason, we  say that 1/0 is undefined.

A simple analogy would also let us realize that allowing division by 0 will violate an important property of real numbers.  For example 8/4 = 2 because 2 x 4 = 8.  Assuming division of 0 is allowed. If 5/0 = n, then n x 0 = 5.  Now, that violates the property of a real number that any number multiplied by 0 is equal to 0.

Since division by 0 yields an answer which is not defined, the said operation is not allowed.

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An extensive explanation about the Epsilon-Delta definition of limits

One of the most important topics in elementary calculus is the \epsilon-\delta definition of limits.  The definition says that the \lim_{x \to a} f(x)= L if and only if, for all \epsilon > 0, there exists a \delta >0 such that if |x-a| < \delta , then |f(x) - L |<\epsilon. In this article, we are going to discuss what this definition means. Readers of this article must have knowledge about elementary calculus and the concept of limits.

Review of Limit Basics

Consider the function f(x) = 2x. We have learned from elementary calculus that \lim_{x \to 2} 2x = 4.  Aside from algebraic computation, this is evident from the color-coded graph and the table shown in Figure 1.  The yellow arrows in the graph and the values in the yellow cells in the table indicate that as the value of x approaches 2 from the left of the x-axis, the value of f(x) approaches 4 from below of the y-axis. On the other hand, the red arrows in the graph and the values in the red cells in the table indicate that as the value of x approaches 2 on from the right of the x-axis, the value of f(x) approaches 4 from above of the y-axis.

Figure 1 – The table and the graph showing the value of f(x) as x approaches 2 from both sides.

From the above discussion, it is noteworthy to mention three things:

  1. We can get f(x) as close to 4 as we please by choosing an x sufficiently close to 2.  For example, I can set x to 1.9999 \cdots99 (with 100 nines) to get an f(x) very close to 4, which is 3.9999 \cdots 98 (99 nines).
  2. No matter how small is the distance of x from 2, a distance less than it may still be chosen. For example, if we choose the point which is very close to 2, say a point with coordinate 1.9999 \cdots 99 with (100 nines), we can still choose a value closer than this to 2.  For instance, we can choose 1.9999 \cdots 99 with 101 nines. This can be repeated for every chosen distance.
  3. Although x can be very very close to 2, it does not necessarily mean that x equals 2.

Now we go back to the definition of limits. In a specific example, the limit definition states that the \lim_{x \to 2} 2x = 4 if (and only if) for all distance (denoted by the Greek letter \epsilon) from 4  along the y-axis (directly above or below 4) – no matter how small – we can always find a certain distance (denoted by \delta) from 2 along the x-axis (left or right of 2) such that if x is between 2 - \delta and 2 + \delta,  then f(x) would lie between 4 - \epsilon and 4 + \epsilon.

To give you a more concrete example, suppose we want the distance \epsilon from 4, which is our limit, to be 0.1 then the interval of our f(x) is (4 -0.1, 4+ 0.1) = (3.9,4.1). The definition of limit says that given a distance \epsilon = 0.1, we can find a distance \delta in the x-axis such that if x is between 2 - \delta and 2 + \delta, we are sure that f(x) is between 3.9 and 4.1. We do not know the value of \delta yet, but we will calculate it later.

Figure 2 – The epsilon-delta definition given epsilon = 0.1.

In Figure 2, x is between 2 - \delta and 2 + \delta or 2 - \delta < x < 2 + \delta. Subtracting 2 from all terms of the inequality, we have - \delta < x - 2 < \delta. If you recall the definition of absolute value, this is precisely the same as |x - 2| < \delta. The comparison among the notations is in Table 1.

Using the notations in the table, we can conclude that the following statements are equivalent:

  1. Words: Given \epsilon = 0.1, we can find a \delta such that if x is between 2 - \delta, 2 + \delta, then f(x) is between 3.9 and 4.1.
  2. Set Notation: Given \epsilon = 0.1, we can find a \delta such that if x \in (2 -\delta, 2 + \delta), then f(x) \in (3.9,4.1).
  3. Relational Operator: Given \epsilon = 0.1, we can find a \delta such that if 2 - \delta < x < 2 + \delta, then 3.9< f(x) < 4 .1.
  4. Absolute Value: Given \epsilon = 0.1, we can find a \delta such that if |x-2| < \delta, then |f(x) - 4| < 0.1.

We have discussed that we can get f(x) as close to 4 as we please

 by choosing an x sufficiently close to 2.  This is equivalent to choosing an extremely small \epsilon, no matter how small, as long as \epsilon>0. Our next task is to find the \delta that corresponds to that \epsilon.

Applying this definition to our example, we can say the \lim_{x \to 2} 2x = 4 if and only if, given \epsilon > 0 (any small distance above and below 4), we can find a \delta > 0 (any distance from x to the left and right of 2) such that if |x - 2| < \delta, then |f(x) - 4| < \epsilon.

The Definition of a Limit of a Function

Now, notice that 4 is the limit of the function as x approaches 2. If we let the limit of a function be equal to L and a be the fixed value that x approaches, then we can say that \lim_{x \to a} f(x) = L if and only if, for any \epsilon>0 (any small distance above and below L), we can find a \delta>0 (any small distance from to the left and to the right of a) such that if |x - a| < \delta then, |f(x) - L| < \epsilon. And that is precisely, the definition of limits that we have stated in the first paragraph of this article.

Figure 3 – The epsilon-delta definition given any epsilon.

In mathematics, the phrase “for any” is the same as “for all” and is denoted by the symbol \forall. In addition, the phrase “we can find” is also the same as “there exists” and is denoted by the symbol \exists. So, rephrasing the definition above, we have \lim_{x \to a} f(x) = L if and only if, \forall \epsilon > 0, \exists \delta >0, such that if |x - a| < \delta then, |f(x) - L| < \epsilon. A much shorter version of this definition is the phrase \lim_{x \to a} f(x) = L \iff, \forall \epsilon > 0, \exists \delta >0, such that |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon. The symbol \iff stands for if and only if and the symbol \Rightarrow is similiar to if-then. If P and Q are statements, the statement P \Rightarrow Q is the same as the statement of the form “If P then Q“.

Finding a specific delta

We said that given any positive \epsilon, we can find a specific \delta, no matter how small our \epsilon is. So let us try our first specific value \epsilon = 0.1.

From the definition, we have \lim_{x \to 2} 2x = 4 if and only if, given \epsilon> 0 (any small distance above and below 4), \exists \delta >0 such that if |x -2| < \delta then, |f(x) - 4| < \epsilon.

Now |f(x) - 4| = |2x - 4| < \epsilon. This implies that
|2||x-2| < \epsilon which implies that 2|x - 2| <0.1. Simplifying, we have |x- 2| < 0.05. This means that our x should be between 1.95 and 2.05 to be sure that our f(x) is between 3.9 and 4.1. This is shown in Figure 4.

Figure 4 – The table showing some of the values of epsilon and delta satisfying the definition of limit of 2x as x approaches 2.

Now, let \epsilon = 0.05. This means that our interval is (3.95,4.05). Now |f(x) - 4| = |2x - 4| < 0.05. Thus, |2||x -2| < \epsilon which implies that 2|x - 2| <0.05. Solving, we have |x - 2| < 0.025. This means that our x should be between 1.975 and 2.025 to be sure that our f(x) is between 3.95 and 4.05. There are only two examples above, but the definition tells us that we can choose any \epsilon > 0 so let us generalize our statement by doing so.

Now |f(x) - 4| = |2x - 4| < \epsilon. This results to |2||x -2| < \epsilon which implies that 2|x - 2| <\epsilon. Solving, we have |x - 2| < \epsilon/2. From the condition above, |x - 2| < \delta so we can let \delta = \epsilon/2.

This means given any \epsilon, we just let our \delta equal to \epsilon/2 and we are sure that if x is between 2-\delta/2 and 2+\delta to be sure that our f(x) is between 4 - \epsilon and 4 - \epsilon.

In the next calculus post, we are going to discuss the strategies on how to get  \delta given an arbitrary \epsilon value, so keep posted.

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