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# Identifying Rational Numbers

Alignments to Content Standards: 8.NS.A.1

Decide whether each of the following numbers is rational or irrational. If it is rational, explain how you know.

1. $0.33\overline{3}$
2. $\sqrt{4}$
3. $\sqrt{2}= 1.414213…$
4. $1.414213$
5. $\pi = 3.141592…$
6. $11$
7. $\frac17=0.\overline{142857}$
8. $12.34565656\overline{56}$

## IM Commentary

The task assumes that students are able to express a given repeating decimal as a fraction. Teachers looking for a task to fill in this background knowledge could consider the related task "8.NS Converting Decimal Representations of Rational Numbers to Fraction Representations ."

There is a lot of interesting mathematics behind deciding questions about irrationality. There are a variety of arguments demonstrating that $\sqrt{2}$ is irrational (some of which would be quite accessible to a motivated middle school student), the first of which were discovered somewhere around the 5th century BC. And yet the irrationality of $\pi$ was not proven until 1761, over two millenia later! Students who complete the task will probably be very close to being able to articulate the statement that a number is rational if and only if its decimal expansion is eventually periodic, in which case they could be posed problems like showing that the number $$0.123456789101112131415161718192021....$$ is irrational. Note that even understanding the statement that $\sqrt{2}$ equals $1.414213…$ is non-trivial, and partly addresses the part of the standard that says “Understand informally that every number has a decimal expansion."

## Solution

1. Since $$0.33\overline{3} = \frac13$$ $0.33\overline{3}$ is a rational number.
2. Since $$\sqrt{4} = 2 =\frac21$$ $\sqrt{4}$ is a rational number.
3. $\sqrt{2} = 1.414213…$ is not rational. In eighth grade most students know that the square root of a prime number is irrational as a "fact," but few 8th grade students will be able to prove it. There are arguments that 8th graders can understand if they are interested.
4. Since $$1.414213 = \frac{1414213}{100000},$$ $1.414213$ is a rational number.

5. $\pi = 3.141592…$ is not rational. In eighth grade most students know that $\pi$ is irrational as a "fact." The proof of this is quite sophisticated.

6. Since $$11=\frac{11}{1}$$ 11 it is rational.

7. $\frac17=0.\overline{142857}$ is already written in a way that makes it clear it is a rational number, although some students might say it is irrational, possibly because the repeating part of the decimal is longer than many familiar repeating decimals (like $\frac13$).

8. We have $$12.34565656\overline{56} = 12.34+.00\overline{56}=\frac{1234}{100}+\frac{56}{9900}=\frac{1234\cdot 99+56}{9900}=\frac{122222}{9900},$$ which is certainly rational.

#### Jason says:

over 6 years

The solution for part (a) doesn't indicate the reasoning - maybe the solution could refer people to the task called "Converting Decimal Representations of Rational Numbers to Fraction Representations," where the reasoning is given.

Is part (f) somewhat more advanced than grade 8? I'm not saying it's terribly out of place here, but it does seem more in line with high school standard N-RN.3: "Explain why...the product of a nonzero rational number and an irrational number is irrational." Part (d) is more or less an instance of this, the product in question being $\frac{1}{2}\times\pi$.

Likewise, the solution for part (i) refers to one of the principles called out in N-RN.3. Instead of appealing to this principle, perhaps one might simply observe that the two fractions on the right-hand side could be added to form a single fraction by finding a common denominator, showing that the number is rational.

Or perhaps parts (f) and (i) might be good illustrations for N-RN.3 (that standard doesn't currently have any tasks).

Note typo in solution to part (d).

#### Cam says:

over 6 years

Great, thanks. I've updated the commentary to make expectations more clear and add the reference, fixed the typo in (d), and re-written the solution to part (i) (which was actually incorrect) to remove the language reflective of the N-RN standards. I agree that (f) is more advanced than the rest, though perhaps manageably more so -- I've left it alone pending further discussion.

#### Jason says:

over 6 years

Thanks Cam.

The fix to the Commentary says, "The task assumes that students know the relationship between rational numbers and repeating decimal expansions (i.e., that they are able to determine that, e.g., $\frac{1}{3} = 0.333\dots$)."

I think I get what you mean, but I'm not sure that what comes after the i.e. is really equivalent to what comes before it. In any case, I think it would be clearer to say, simply, "The task assumes that students are able to express a given repeating decimal as a fraction." This is the particular skill being used in part (a).

(The existing language of "Determining that $\frac{1}{3} = 0.333\dots$" sounds like it could refer to the reverse process; that is, a student could determine that $\frac{1}{3} = 0.333\dots$ by doing a long division problem on $\frac{1}{3}$. But expressing a given fraction as a repeating decimal by long division is a skill not required for part (a), and also not sufficient, unless say the student is trying fractions at random and happens upon $\frac{1}{3}$.)

As for part (f), we can see if others weigh in. My argument is twofold: not only that it's more advanced (in which case we might well leave it in), but that it's more advanced and forms the subject of a high school standard, in which case for alignment purposes it really ought to go.

Best, J

#### Kristin says:

I agree that your wording in the commentary is more succinct and clear and changed it to what you suggest. I had also struggled with whether the case of $\frac{\pi}{2}$ was appropriate for 8th grade, and after I came across this conversation I went ahead and removed it. I'll start a task for N-RN-3.