Download Math 285 Logic Homework Problems

Math 285 Logic/Proof Homework Problems
1) Write the a) converse b) contrapostive for each of the following:
i) If x is even, then x+1 is odd
ii) If a function is differentiable, then it is continuous.
iii) If you have a college degree, then you can work here.
2)
Write each biconditional statement as two conditional statements:
i)
ii)
A function is even iff f(-x) = f(x)
A system of linear equations is consistent iff the dimension of the column space is n.
3)
Give, if possible, an examples of a true conditional statement for which a) the converse is false b) the
converse is true c) the contrapostive is false.
4) Give a counterexample for each of the following:
i)
If a function is continuous, then it is differentiable.
ii) If x  4 , then x  2
5) Use contrapositive/contradiction to prove each of the following:
2
i)
ii)
iii)
n  N . If 3n  1 is even, then n is odd.
2
Let n  N . If n is even, then n is even.
Show 2 is irrational.
Let
Answer Key
1) i) Converse: If x+1 is odd, then x is even.
Contrapostive: If x+1 is not odd, then x is not even (or if x+1 is even, then x is odd)
ii) Converse: If a function is continuous, then it is differentiable
Contrapostive: If a function is not continuous, then it is not differentiable.
iii)
Converse: If you can work here, then you have a college degree.
Contrapostive: If you cannot work here, then you don’t have a college degree.
2) i) If a function is even then f(-x) = f(x) and If a function satisfies f(-x) = f(x), then it is even.
ii) If a system of linear equations is consistent then the dimension of the column space is n, and if the
dimension of the column space is n, then the system of linear equations is consistent .
3) a) part II of question 1 b) part I of question 1. c) not possible,
4)i)
f ( x)  x
is continuous, but it is not differentiable at x = 0. ii) x = -3.
5)
a) Suppose n is even. By definition of an even number, n=2k for some k. Then
3n  1  3(2k )  1  2(3k )  1 .
Therefore,
3n  1 is odd.
b) Prove by contraposition. Suppose n is odd. By definition of an odd number, n=2k+1 for some k.
Then
n2  (2k  1)2  4k 2  4k  1  2(2k 2  2k )  1
is odd.