Difference between revisions of "1963 AHSME Problems/Problem 26"

(Created page with "== Problem == Consider the statements: <math> \textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\...")
 
(Solution)
 
Line 23: Line 23:
 
Statement <math>4</math> states that <math>p</math> is false and <math>q</math> is true.  In this case, <math>p \rightarrow q</math> is true - your conclusion can be true even if your premise is false.  And, since <math>r</math> is also true from statement <math>4</math>, this means <math>(p \rightarrow q) \rightarrow r</math> is true.  Thus, statement <math>4</math> implies the truth of the given statement.
 
Statement <math>4</math> states that <math>p</math> is false and <math>q</math> is true.  In this case, <math>p \rightarrow q</math> is true - your conclusion can be true even if your premise is false.  And, since <math>r</math> is also true from statement <math>4</math>, this means <math>(p \rightarrow q) \rightarrow r</math> is true.  Thus, statement <math>4</math> implies the truth of the given statement.
  
All four statements imply the truth of the given statement, so the answer is <math></math>\boxed{\textbf{(E)}}$
+
All four statements imply the truth of the given statement, so the answer is <math>\boxed{\textbf{(E)}}</math>
 
 
  
 
==See Also==
 
==See Also==

Latest revision as of 17:32, 22 July 2019

Problem

Consider the statements:

$\textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\wedge\sim q\text{ }\wedge\sim r\qquad\textbf{(4)}\ \sim p\text{ }\wedge q\text{ }\wedge r$

where $p,q$, and $r$ are propositions. How many of these imply the truth of $(p\rightarrow q)\rightarrow r$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution

Statement $1$ states that $p$ is true and $q$ is false. Therefore, $p \rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false. This means that $(p \rightarrow q) \rightarrow X$, where $X$ is any logical statement (or series of logical statements) must be true - if your premise is false, then the implication is automatically true. So statement $1$ implies the truth of the given statement.

Statement $3$ similarly has $p$ as true and $q$ is false, so it also implies the truth of the given statement.

Statement $2$ states that $p$ and $q$ are both false. This in turn means that $p \rightarrow q$ is true. Since $r$ is also true from statement $2$, this means that $(p \rightarrow q) \rightarrow r$ is true, since $T \rightarrow T$ is $T$. Thus statement $2$ implies the truth of the given statement.

Statement $4$ states that $p$ is false and $q$ is true. In this case, $p \rightarrow q$ is true - your conclusion can be true even if your premise is false. And, since $r$ is also true from statement $4$, this means $(p \rightarrow q) \rightarrow r$ is true. Thus, statement $4$ implies the truth of the given statement.

All four statements imply the truth of the given statement, so the answer is $\boxed{\textbf{(E)}}$

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS