How to graph an equation with a number line - PSAT Math
Card 1 of 7
If 0 < n < 1, then which of the following is the smallest?
If 0 < n < 1, then which of the following is the smallest?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
If 0 < n < 1, then which of the following is the smallest?
If 0 < n < 1, then which of the following is the smallest?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
If 0 < n < 1, then which of the following is the smallest?
If 0 < n < 1, then which of the following is the smallest?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
If 0 < n < 1, then which of the following is the smallest?
If 0 < n < 1, then which of the following is the smallest?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
If 0 < n < 1, then which of the following is the smallest?
If 0 < n < 1, then which of the following is the smallest?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
If 0 < n < 1, then which of the following is the smallest?
If 0 < n < 1, then which of the following is the smallest?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
If 0 < n < 1, then which of the following is the smallest?
If 0 < n < 1, then which of the following is the smallest?
Tap to reveal answer
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.
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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