1974 AHSME Problems/Problem 13

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 $\boxed{\text{D}}$

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1974 AHSME Problems/Problem 13

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