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Math 121 - Homework 1
Due Wednesday, Sept 7
Before starting this homework, read “Mathematical hygiene” from the files section of the canvas
site. You may also want to read Cheng’s proof guide (also on the canvas site), but it is not required.
1. Rewrite each of the following sentences in the form “If P , then Q.”
(a) The product of an odd integer and an even integer is even.
(b) The square of an odd integer is odd.
(c) The square of a prime number is not prime.
2. Write the negation of the following statements without just adding the word “not.” Assume
that x is a real number.
(a) All cows are brown.
(b) x ≤ 2.
(c) All cows are brown and eat grass.
(d) x is a prime number greater than 12.
(e) There exists a cow that is black and doesn’t eat grass.
3. (a) Come up with two statements, P and Q such that “If P , then Q” is true but “If Q, then
P ” is false.
(b) Using your statements from (a), write out “If ¬Q, then ¬P .” (“If ¬Q, then ¬P ” is called
the contrapositive of “If P , then Q.” Notice that these two statements are equivalent! )
4. A direct proof of a statement of the form “If P , then Q” starts by assuming that P is true,
and then, after a series of deductions, ends with the conclusion that Q is true. For example,
consider the statement “Let a be an arbitrary element of a field. Then a · 0 = 0.” (Theorem
C.2(a) in the book.)
Here is a direct proof:
Assume a is an arbitrary element of a field. Since 0 is an additive identity, we know that
0 + 0 = 0. By distributivity,
0 + a · 0 = a · 0 = a · (0 + 0) = a · 0 + a · 0.
By the cancellation law of addition (proved in class), we can subtract a · 0 from each side, and
we get
0 = a · 0,
as desired.
Show, using a direct proof, that if x is an element of a field, then −(−x) = x. Make sure you
mention which field axioms you are using.
5. A proof by contradiction starts by assuming the statement is false, and ends, after a series
of deductions with something we know to be false. For example, consider the statement
“There are infinitely many prime numbers.”
Here is a proof by contradiction:
Assume there are only finitely many primes p1 , p2 , . . . , pn . Consider the integer p = p1 p2 · · · pn +
1. p is greater than all the primes, so it can’t be prime itself. Thus, it must be divisible by
some prime pi . p/pi = (p1 p2 · · · pn )/pi + 1/p1 . p1 p2 · · · pn )/pi is an integer but 1/p1 is not.
Thus, p/pi cannot be an integer, which contradicts the fact that pi divides p. Therefore, there
must be infinitely many primes.
√
Prove by contradiction that 2 is not a rational number.
6. Let F be a field, and let a and b be elements of F . Show that if a · b = 0, then a = 0 or b = 0
(or both).
7. Let F be a field, and let a and b be nonzero elements of F . Show that there is a unique
element c ∈ F such that a · c = b.
8. Show that Q is the smallest field contained in R that contains Z (i.e. every other field
contained in R that contains Z must also contain Q).
9. Let F (along with the operations + and ·) be a field. Consider the set F 2 ( i.e. F × F ). Show
that F 2 equipped with the operations (a, b)+(c, d) = (a+c, b+d) and (a, b)∗(c, d) = (a·c, b·d)
is not a field.
10. Let X = {a, b, c, d}. Come up with an addition table and a multiplication table for X so that
X a field. Which element is the additive identity? Multiplicative identity? List the additive
and multiplicative inverses for each (nonzero) element.