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. | ||
+ | |||
+ | ==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: | ||
+ | <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. | ||
+ | |||
+ | <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! | ||
+ | |||
+ | <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! | ||
+ | |||
+ | <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. | ||
+ | |||
+ | <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. | ||
+ | |||
+ | The statement that makes the most sense is <math> \boxed{\text{D}} </math> | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1974|num-b=12|num-a=14}} | {{AHSME box|year=1974|num-b=12|num-a=14}} | ||
+ | [[Category:Logic Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:04, 27 June 2024
Contents
Problem
Which of the following is equivalent to "If P is true, then Q is false."?
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."
The fact that a statement's contrapositive is logically equivalent to it can easily be seen from a venn diagram arguement.
From this venn diagram, clearly "If , then
." is true. However, since
is fully contained in
, the statement "If not
, then not
." 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:
---> If the dog is 100% blue or the dog is not 100% red. This statement makes no sense.
---> 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!
---> If the dog is not 100% blue, then the dog is 100% red. Again, this fails for other colors of dogs. This is dogscrimination!
---> If the dog is 100% red, then it is not 100% blue. This statement makes logical sense, and is correct.
---> 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.