Algebra 1 PARCC question: rational-irrational

Correct answers: a is rational, b is irrational, and the product of a and b is irrational.

PARCC evidence statement(s) tested: HS.C.2.1:

Base explanations/reasoning on the properties of rational and irrational numbers.

For rational solutions, exact values are required. For irrational solutions, exact or decimal approximations may be required. Simplifying or rewriting radicals is not required; however, students will not be penalized if they simplify the radicals correctly.

The evidence statement above references Math Practice 3 in the Common Core: Construct viable arguments and critique the reasoning of others. Mathematically proficient students understand and use stated assumptions, definitions, and previously established results in constructing arguments. They make conjectures and build a logical progression of statements to explore the truth of their conjectures. They are able to analyze situations by breaking them into cases, and can recognize and use counterexamples. They justify their conclusions, communicate them to others, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. …

The question tests students’ understanding of the high school Common Core algebra standard HSN.RN.B.3, found under high school algebra (Number and quantity: the real number system), which states, under “use properties of rational and irrational numbers,” that they should be able to “explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational.”

Construct an argument to justify your answers.

To justify that a is rational, you need only state that it can be represented as an integer divided by another integer. That’s the definition of a rational number.

To justify that b is irrational, you need only state that it cannot be represented as an integer divided by another integer. It is a non-repeating, non-terminating decimal. Those are never rational.

When it comes to the product of a and b, you could show that the actual product of these two numbers, each truncated at the 12th decimal place, is a non-repeating, non-terminating decimal that can’t be represented as an integer divided by another integer.

That doesn’t look like it’s ever going to end or repeat. But a better way to do it (although no algebra 1 test-taker on last year’s PARCC sample student responses did it this way) would be to show that, in the general case, the product of a nonzero rational number, such as a, and an irrational number is an irrational number.

(Anything times 0 is 0, which is a rational number, so that’s why we have to be clear that we’re talking about the product of the irrational number with a rational number that is not 0.)

On with the reasoning. If we assume R is a rational number not equal to 0 and S is an irrational number, then R can be represented as some integer divided by another integer. Let’s use p and q for our argument.

The irrational number S on the other hand cannot be represented as one integer divided by another number, because no irrational number can be represented that way, by definition.

Multiply R × S and we get

Now the trick. Ask yourself: Is there any possibility that the product can be a rational number? Whether there is or isn’t, we’re going to assert that it can be rational. That would mean it could be represented as one integer divided by another integer, let’s say u/v.

Solve that for S and we get

That’s a problem, since the product of two integers, u and q, is an integer, as is the product of v and p. That means S is equal to one integer divided by another integer, which contradicts our initial assumption that S was an irrational number.

This is called a proof by contradiction, and it means that, in general, any rational number not equal to 0 (note that if v or p = 0, the fraction above is undefined) times an irrational number will result in an irrational number as the product.