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Difference between revisions of "2000 AMC 8 Problems/Problem 2"
 
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The number <math>0</math> has no reciprocal, and <math>1</math> and <math>-1</math> are their own reciprocals. This leaves only <math>2</math> and <math>-2</math>. The reciprocal of <math>2</math> is <math>1/2</math>, but <math>2</math> is not less than <math>1/2</math>. The reciprocal of <math>-2</math> is <math>-1/2</math>, and <math>-2</math> is less than<math> -1/2</math>, so it is <math>\boxed{A}</math>
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==Problem==
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Which of these numbers is less than its reciprocal?
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<math>\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2</math>
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==Solution==
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===Solution 1===
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The number <math>0</math> has no reciprocal, and <math>1</math> and <math>-1</math> are their own reciprocals. This leaves only <math>2</math> and <math>-2</math>. The reciprocal of <math>2</math> is <math>1/2</math>, but <math>2</math> is not less than <math>1/2</math>. The reciprocal of <math>-2</math> is <math>-1/2</math>, and <math>-2</math> is less than<math> -1/2</math>, so it is <math>\boxed{A}</math>.
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===Solution 2===
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The statement "a number is less than its reciprocal" can be translated as <math>x < \frac{1}{x}</math>.
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Multiplication by <math>x</math> can be done if you do it in three parts:  <math>x>0</math>, <math>x=0</math>, and <math>x<0</math>.  You have to be careful about the direction of the inequality, as you do not know the sign of <math>x</math>.
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If <math>x>0</math>, the sign of the inequality remains the same.  Thus, we have <math>x^2 < 1</math> when <math>x > 0</math>.  This leads to <math>0 < x < 1</math>.
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If <math>x=0</math>, the inequality <math>x < \frac{1}{x}</math> is undefined.
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If <math>x<0</math>, the sign of the inequality must be switched.  Thus, we have <math>x^2 > 1</math> when <math>x < 0</math>.  This leads to <math>x < -1</math>.
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Putting the solutions together, we have <math>x<-1</math> or <math>0 < x < 1</math>, or in interval notation, <math>(-\infty, -1) \cup(0, 1)</math>.  The only answer in that range is <math>\boxed {\text{(A)}\ -2}</math>
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===Solution 3===
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Starting again with <math>x < \frac{1}{x}</math>, we avoid multiplication by <math>x</math>.  Instead, move everything to the left, and find a common denominator:
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<math>x < \frac{1}{x}</math>
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<math>x - \frac{1}{x} < 0</math>
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<math>\frac{x^2 - 1}{x} < 0</math>
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<math>\frac{(x+1)(x-1)}{x} < 0</math>
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Divide this expression at <math>x=-1</math>, <math>x=0</math>, and <math>x=1</math>, as those are the three points where the expression on the left will "change sign".
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If <math>x<-1</math>, all three of those terms will be negative, and the inequality is true.  Therefore, <math>(-\infty, -1)</math> is part of our solution set.
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If <math>-1 < x < 0</math>, the <math>(x+1)</math> term will become positive, but the other two terms remain negative.  Thus, there are no solutions in this region.
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If <math>0 < x < 1</math>, then both <math>(x+1)</math> and <math>x</math> are positive, while <math>(x-1)</math> remains negative.  Thus, the entire region <math>(0, -1)</math> is part of the solution set.
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If <math>1 < x</math>, then all three terms are positive, and there are no solutions.
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At all three "boundary points", the function is either <math>0</math> or undefined.  Therefore, the entire solution set is <math>(-\infty, -1) \cup (0, 1)</math>, and the only option in that region is <math>x=-2</math>, leading to <math>\boxed{A}</math>.
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==Solution 4==
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We can find out all of their reciprocals. Now we compare and see that the answer is <math>\boxed{A}</math>
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==Solution 5==
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Look at each number. Notice that the number must be negative. The number cannot be -1, 0, 1, ... .  -2 is all that is left
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==See Also==
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{{AMC8 box|year=2000|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 12:36, 10 July 2019

Problem

Which of these numbers is less than its reciprocal?

$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution

Solution 1

The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$. The reciprocal of $2$ is $1/2$, but $2$ is not less than $1/2$. The reciprocal of $-2$ is $-1/2$, and $-2$ is less than$-1/2$, so it is $\boxed{A}$.

Solution 2

The statement "a number is less than its reciprocal" can be translated as $x < \frac{1}{x}$.

Multiplication by $x$ can be done if you do it in three parts: $x>0$, $x=0$, and $x<0$. You have to be careful about the direction of the inequality, as you do not know the sign of $x$.

If $x>0$, the sign of the inequality remains the same. Thus, we have $x^2 < 1$ when $x > 0$. This leads to $0 < x < 1$.

If $x=0$, the inequality $x < \frac{1}{x}$ is undefined.

If $x<0$, the sign of the inequality must be switched. Thus, we have $x^2 > 1$ when $x < 0$. This leads to $x < -1$.

Putting the solutions together, we have $x<-1$ or $0 < x < 1$, or in interval notation, $(-\infty, -1) \cup(0, 1)$. The only answer in that range is $\boxed {\text{(A)}\ -2}$

Solution 3

Starting again with $x < \frac{1}{x}$, we avoid multiplication by $x$. Instead, move everything to the left, and find a common denominator:

$x < \frac{1}{x}$

$x - \frac{1}{x} < 0$

$\frac{x^2 - 1}{x} < 0$

$\frac{(x+1)(x-1)}{x} < 0$

Divide this expression at $x=-1$, $x=0$, and $x=1$, as those are the three points where the expression on the left will "change sign".

If $x<-1$, all three of those terms will be negative, and the inequality is true. Therefore, $(-\infty, -1)$ is part of our solution set.

If $-1 < x < 0$, the $(x+1)$ term will become positive, but the other two terms remain negative. Thus, there are no solutions in this region.

If $0 < x < 1$, then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative. Thus, the entire region $(0, -1)$ is part of the solution set.

If $1 < x$, then all three terms are positive, and there are no solutions.

At all three "boundary points", the function is either $0$ or undefined. Therefore, the entire solution set is $(-\infty, -1) \cup (0, 1)$, and the only option in that region is $x=-2$, leading to $\boxed{A}$.

Solution 4

We can find out all of their reciprocals. Now we compare and see that the answer is $\boxed{A}$

Solution 5

Look at each number. Notice that the number must be negative. The number cannot be -1, 0, 1, ... . -2 is all that is left

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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