PSAT Math : How to graph an equation with a number line
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Example Question #51 : Integers
If 0 < n < 1, then which of the following is the smallest?
1/n2
–1/n3
n3 – 1
n2
–1/n
–1/n3
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 n2 is positive, because any number squared is positive. Similarly, 1/n2 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/n3 will also be negative.
The last choice is n3 – 1. We are told that 0 < n < 1. Because n is a positive value less than one, we know that 0 < n3 < n2 < n < 1. In other words, n3 will be a small positive value, but it will still be less than one. Thus, because n3 < 1, if we subtract 1 from both sides, we see that n3 – 1 < 0. Therefore, n3 - 1 is a negative value.
All negative numbers are less than positive numbers. Thus, we can eliminate n2 and 1/n2, which are both positive. We are left with –1/n, –1/n3, and n3 – 1.
Let us compare –1/n and –1/n3. First, let us assume that –1/n < –1/n3.
–1/n < –1/n3
Multiply both sides by n3. We don't need to switch the signs because n3 is positive.
–n2 < –1
Multiply both sides by –1.
n2 > 1.
We know that n2 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/n3. Therefore, –1/n3 must be smaller.
Finally, let's compare –1/n3 and n3 – 1. Let us assume that –1/n3 < n3 – 1.
–1/n3 < n3 – 1
Multiply both sides by n3.
–1 < n6 – n3
Add one to both sides.
n6 – n3 + 1 > 0. If this inequality is true, then it will be true that –1/n3 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 n6 – n3 + 1 > 0, –1/n3 is indeed the smallest number. We can verify this by trying more values of n.
The answer is –1/n3.
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