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Difference between revisions of "1974 AHSME Problems/Problem 13"
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From this venn diagram, clearly "If <math> p </math>, then <math> q </math>." is true. However, since <math> p </math> is fully contained in <math> q </math>, the statement "If not <math> q </math>, then not <math> p </math>." is also true, and so a statement and its contrapositive are equivalent.
 
From this venn diagram, clearly "If <math> p </math>, then <math> q </math>." is true. However, since <math> p </math> is fully contained in <math> q </math>, the statement "If not <math> q </math>, then not <math> p </math>." is also true, and so a statement and its contrapositive are equivalent.
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==Solution==
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Let's consider the following case of dogs. If p = true then the dog is 100% blue. If q = true then the dog is 100% red.
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This is a valid solution to if P is true, then Q is false, or if the dog is 100% blue then the dog is not 100% red.
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Now let's look at the options:
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<math> \mathrm{(A)\ } \text{``P is true or Q is false."} \qquad </math> ---> If the dog is 100% blue or the dog is not 100% red. This statement makes no sense.
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<math> \mathrm{(B) \ }\text{``If Q is false then P is true."} \qquad </math> ---> If the dog is not 100% red, then the dog is 100% blue. This does not make sense, as the dog could be other colors, like white, or black, or yellow. My dog is brown!
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<math> \mathrm{(C) \  } \text{``If P is false then Q is true."} \qquad </math> ---> If the dog is not 100% blue, then the dog is 100% red. Again, this fails for other colors of dogs. This is dogscrimination!
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<math> \mathrm{(D) \  } \text{``If Q is true then P is false."} \qquad </math> ---> If the dog is 100% red, then it is not 100% blue. This statement makes logical sense, and is correct.
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<math>  \mathrm{(E) \  }\text{``If Q is true then P is true."} \qquad  </math> --->  If the dog is 100% red, then the dog is 100% blue. This does not make sense, because we know the color of the dog, a dog cannot be two colors at once. Actually in quantum mechanics maybe, but not for this AHSME problem.
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The statement that makes the most sense is <math> \boxed{\text{D}} </math>
  
 
==See Also==
 
==See Also==

Latest revision as of 10:04, 27 June 2024

Problem

Which of the following is equivalent to "If P is true, then Q is false."?

$\mathrm{(A)\ } \text{``P is true or Q is false."} \qquad$

$\mathrm{(B) \ }\text{``If Q is false then P is true."} \qquad$

$\mathrm{(C) \ } \text{``If P is false then Q is true."} \qquad$

$\mathrm{(D) \ } \text{``If Q is true then P is false."} \qquad$

$\mathrm{(E) \ }\text{``If Q is true then P is true."} \qquad$

Solution

Remember that a statement is logically equivalent to its contrapositive, which is formed by first negating the hypothesis and conclusion and then switching them. In this case, the contrapositive of "If P is true, then Q is false." is "If Q is true, then P is false." $\boxed{\text{D}}$

The fact that a statement's contrapositive is logically equivalent to it can easily be seen from a venn diagram arguement.

[asy] draw(circle((0,0),3)); draw(circle((-1,1),1)); label("$q$",(1,-1)); label("$p$",(-1,1)); [/asy]

From this venn diagram, clearly "If $p$, then $q$." is true. However, since $p$ is fully contained in $q$, the statement "If not $q$, then not $p$." is also true, and so a statement and its contrapositive are equivalent.

Solution

Let's consider the following case of dogs. If p = true then the dog is 100% blue. If q = true then the dog is 100% red.

This is a valid solution to if P is true, then Q is false, or if the dog is 100% blue then the dog is not 100% red.

Now let's look at the options: $\mathrm{(A)\ } \text{``P is true or Q is false."} \qquad$ ---> If the dog is 100% blue or the dog is not 100% red. This statement makes no sense.

$\mathrm{(B) \ }\text{``If Q is false then P is true."} \qquad$ ---> If the dog is not 100% red, then the dog is 100% blue. This does not make sense, as the dog could be other colors, like white, or black, or yellow. My dog is brown!

$\mathrm{(C) \ } \text{``If P is false then Q is true."} \qquad$ ---> If the dog is not 100% blue, then the dog is 100% red. Again, this fails for other colors of dogs. This is dogscrimination!

$\mathrm{(D) \ } \text{``If Q is true then P is false."} \qquad$ ---> If the dog is 100% red, then it is not 100% blue. This statement makes logical sense, and is correct.

$\mathrm{(E) \ }\text{``If Q is true then P is true."} \qquad$ ---> If the dog is 100% red, then the dog is 100% blue. This does not make sense, because we know the color of the dog, a dog cannot be two colors at once. Actually in quantum mechanics maybe, but not for this AHSME problem.

The statement that makes the most sense is $\boxed{\text{D}}$

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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