### All PSAT Math Resources

## Example Questions

### Example Question #211 : Arithmetic

If 0 < n < 1, then which of the following is the smallest?

**Possible Answers:**

n^{3} – 1

–1/n

–1/n^{3}

1/n^{2}

n^{2}

**Correct answer:**

–1/n^{3}

First, it will help us to determine which of the answer choices are positive and which are negative.

Because n is positive, we know that n^{2} is positive, because any number squared is positive. Similarly, 1/n^{2} is also positive.

Let's look at the answer choice –1/n. This must be negative, because a negative number divided by a positive one will give us a negative number. Similarly, –1/n^{3} will also be negative.

The last choice is n^{3} – 1. We are told that 0 < n < 1. Because n is a positive value less than one, we know that 0 < n^{3} < n^{2} < n < 1. In other words, n^{3} will be a small positive value, but it will still be less than one. Thus, because n^{3} < 1, if we subtract 1 from both sides, we see that n^{3} – 1 < 0. Therefore, n^{3} - 1 is a negative value.

All negative numbers are less than positive numbers. Thus, we can eliminate n^{2} and 1/n^{2}, which are both positive. We are left with –1/n, –1/n^{3}, and n^{3} – 1.^{}

Let us compare –1/n and –1/n^{3}. First, let us assume that –1/n < –1/n^{3}.

–1/n < –1/n^{3}

Multiply both sides by n^{3}. We don't need to switch the signs because n^{3} is positive.

–n^{2 }< –1

Multiply both sides by –1.

n^{2} > 1.

We know that n^{2} will only be bigger than 1 when n > 1 or if n < –1. But we know that 0 < n < 1, so –1/n is not less than –1/n^{3}. Therefore, –1/n^{3 }must be smaller.

Finally, let's compare –1/n^{3} and n^{3} – 1. Let us assume that –1/n^{3} < n^{3} – 1.

–1/n^{3} < n^{3} – 1

Multiply both sides by n^{3}.

–1 < n^{6} – n^{3}

Add one to both sides.

n^{6} – n^{3} + 1 > 0. If this inequality is true, then it will be true that –1/n^{3} is the smallest number.

Here it will be helpful to try some values for n. Let's pick n = 1/2 and see what happens. It will help to use our calculator.

(1/2)^{6} – (1/2)^{3} + 1 = 0.891 > 0.

Therefore, we suspect that because n^{6} – n^{3} + 1 > 0, –1/n^{3 }is indeed the smallest number. We can verify this by trying more values of n.

The answer is –1/n^{3}.

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