# Solutions to Exam 1 (Part I)

```Math 101. Rumbos
Spring 2010
1
Solutions to Exam 1 (Part I)
1. Provide concise answers to the following questions:
(a) A subset, ๐ด, of the real numbers is said to be bounded if there exists a
positive real number, ๐ , such that
โฃ๐โฃ โฉฝ ๐
for all ๐ ∈ ๐ด.
Give the negation of the statement
“๐ด is bounded.”
Answer: The negation of ”๐ด is bounded ” is
For every positive number, ๐ , there exists and an element, ๐, in ๐ด such that โฃ๐โฃ &gt; ๐ .
โก
(b) Let ๐ด denote a subset of the real numbers and ๐ฝ a positive real number.
Give the contrapositive for the following implication:
๐ก ∈ ๐ด ⇒ ๐ก ≤ ๐  − ๐ฝ.
Answer: The contrapositive of “๐ก ∈ ๐ด ⇒ ๐ก ≤ ๐  − ๐ฝ” is
๐ก &gt; ๐  − ๐ฝ ⇒ ๐ก โ∈ ๐ด.
โก
2. Use the ๏ฌeld and order axioms of the real numbers to prove the following.
(a) Let ๐, ๐ ∈ ๐. If ๐๐ = 0, then either ๐ = 0 or ๐ = 0.
Proof: Assume that ๐๐ = 0 and ๐ โ= 0. Then, by Field Axiom (๐น9 ), ๐−1
exists. Multiplying
๐๐ = 0
by ๐−1 on both sides yields
๐−1 (๐๐) = ๐−1 ⋅ 0 = 0,
from which we get that ๐ = 0, where we have used the Field Axioms (๐น7 ),
(๐น9 ) and (๐น10 ).
Math 101. Rumbos
Spring 2010
2
(b) Let ๐ ∈ โ. If ๐ &gt; 1, then ๐ &lt; ๐2 .
Proof: Assume that ๐ &gt; 1. It then follows that ๐ &gt; 0, since 1 &gt; 0. We
also have that ๐ − 1 &gt; 0. Consequently, by the Order Axiom (๐3 ),
๐(๐ − 1) &gt; 0.
Thus, by the distributive property,
๐2 − ๐ &gt; 0,
from which we get that ๐ &lt; ๐2 .
3. Use the completeness axiom of โ to prove that the set of natural numbers is
not bounded above. Deduce, therefore, that for any real number, ๐ฅ, there exists
a natural number, ๐, such that
๐ฅ &lt; ๐.
Proof: Assume by way of contradiction that โ is a bounded above. Then, since
โ is not empty, it follows from the completeness axiom that sup(โ) exists. Thus
there must be ๐ ∈ โ such that
sup(โ) − 1 &lt; ๐.
(1)
It follows from the inequality in (1) that
sup(โ) &lt; ๐ + 1,
where ๐ + 1 ∈ โ. This is a contradiction. Therefore, it must be that case that
โ not bounded above.
Thus, given any real number, ๐ฅ, there must be a natural number, ๐, such that
๐ฅ &lt; ๐.
Otherwise,
๐ โฉฝ ๐ฅ for all ๐ ∈ โ,
which would say that ๐ฅ is an upper bound for โ. But we just proved that โ is
not bounded above.
```