Introduction to Analysis: Fall 2008
Practice problems - Real Number System
MTH 4101/5101
9/3/2008
1. If r is a rational number, (r 6= 0) and x is an irrational number, prove that
r + x and rx are irrational.
Solution: Since the set of all rational numbers, Q is a field, −r is also a
rational number. Now, if r + x is rational, then x = (−r) + (r + x) must also
be a rational number due to the field axioms. But, x is irrational. Hence,
r + x cannot be rational. Similarly, one can show that rx is also rational.
2. Prove that there is no rational number whose square is 12.
Solution Suppose that there is a rational number whose square is 12. Since 4
is a factor for 12, let such a rational number be 2r where r is again a rational
in its lowest terms. Observe
number. This implies, r2 = 3. Suppose r = m
n
that neither m nor n can be even. For, if m is even, say, m = 2k we have
4k 2 = 3n2 implying n is even. (why?) Thus, both m and n are odd. Let
m = 2s + 1 and n = 2t + 1. We then have, (2s + 1)2 = 3(2t + 1)2 . This implies,
2s2 + 2s = 6t2 + 6t + 1 and this is a contradiction.
3. Let E be a non empty subset of an ordered set; Suppose α is a lower bound of
E and β is an upper bound of E. Prove that α ≤ β.
Solution: Let x0 ∈ E. Then we have α ≤ x0 and x0 ≤ β. Transitive property
of order, implies α ≤ β.
4. Let A be a non empty set of real numbers which is bounded below. Let −A be
the set of all numbers −x where x ∈ A. Prove that inf A = - sup(−A).
Solution: Let α =inf A. For any x ∈ A, α ≤ x. This implies, −α ≥ −x for all
−x ∈ −A. This means −α is an upper bound of (−A). Also, If −γ < −α then
−γ cannot be an upper bound of −A. For if it is, then −γ ≥ −x for all
−x ∈ (−A). This implies, γ > α and γ is a lower bound of A which is not
possible since α is the inf A.
5. If a ∈ IR is such that 0 ≤ a < for every > 0. Show that a = 0.
Solution: Let a > 0. Let = a2 . This implies 0 < a < a2 by assumption.
Clearly, this is impossible.
6. Let a, b ∈ IR. Suppose that for every > 0 we have a ≤ b + . Show that a ≤ b.
Solution: Suppose a > b. Then 0 < a − b ≤ . One can see this is impossible,
by choosing = (a−b)
.
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7. Let u be an upper bound of a non empty set S of IR. Prove that u is the
supremum of S if and only if for every > 0 there exists a s ∈ S such that
u − < s .
Solution: Since u is an upper bound of S we have s ≤ u for all s ∈ S.
Suppose u = supS. Then, for any > 0, u − is not an upper bound of S, and
hence there exists an element s ∈ S such that u − < s . Conversely, the
condition given implies that no number less than u can be an upper bound.
Hence, u is the least upper bound of S.
8. If S ⊂ IR contains one of its upper bounds, show that this upper bound of S is
the supremum of S.
Solution: Suppose that u ∈ S is an upper bound of S but not the supremum
of S. Let v 6= u be the supremum of S. Then, x ≤ v for all x ∈ S. Since u ∈ S
we have u < v. There exists a w ∈ S such that u < w. But this contradicts
the fact that u is an upper bound of S.
9. Let A and B be bounded subsets of IR. Show that A ∪ B is bounded. Show
that sup (A ∪ B) = max {sup A, sup B}.
Solution: Let M1 , M2 be real numbers such that |x| ≤ M1 for all x ∈ A and
|y| ≤ M2 for all y ∈ B. Define M = max {M1 , M2 }. Then, |x| ≤ M for all
x ∈ A ∪ B. So, A ∪ B is bounded. Further: Let sup A = u and sup B = v and
w = max{u, v}. For x ∈ A and y ∈ B, we have x ≤ u ≤ w,and y ≤ v ≤ w.
Then, w is an upper bound of A ∪ B. If z is an upper bound of A ∪ B, then z
is an upper bound of A and B. And, u ≤ z and v ≤ z. Hence, w ≤ z.
Therefore, w = (A ∪ B).
10. Let S be a non empty subset of IR that is bounded above. For any a ∈ IR
define a + S = {a + s|s ∈ S}. Prove that sup (a + S) = a+ sup S.
Solution: Let u = sup S, then x ≤ u for all x ∈ S. Then, a + x ≤ a + u.
Thus, a + u is an upper bound for the set a + S. Thus, sup(a + S) ≤ a + u.
Now, if v is any upper bound of the set a + S then a + x ≤ v for all x ∈ S.
Consequently, x ≤ v − a for all x ∈ S so that v − a is an upper bound of S.
Then u = sup S ≤ v − a and so, a + u ≤ v. Since v is any upper bound of
a + S and sup(a + S) is also an upper bound, we have a + u ≤sup (a + S).
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