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Practice problems - Real Number System

1. If r is a rational number , (r ≠ 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
number. This implies, r2 = 3. Suppose r = m/n in its lowest terms. Observe
that neither m nor n can be even. For, if m is even, say, m = 2k we have
4k2 = 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
Solution : Let Then we have and . Transitive property
of order , implies

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, This implies, for all
−x ∈−A. This means −α is an upper bound of (−A). Also, If then
cannot be an upper bound of −A. For if it is, then for all
−x ∈(−A). This implies, and is a lower bound of A which is not
possible since is the inf A.

5. If is such that for every . Show that a = 0.
Solution: Let a > 0. Let . This implies 0 < a < a/2 by assumption.
Clearly, this is impossible.

6. Let Suppose that for every we have Show that
Solution: Suppose a > b. Then One can see this is impossible,
by choosing

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 there exists a such that

Solution: Since u is an upper bound of S we have
Suppose u = supS. Then, for any is not an upper bound of S, and
hence there exists an element such that 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 contains one of its upper bounds, show that this upper bound of S is
the supremum of S.
Solution: Suppose that is an upper bound of S but not the supremum
of S. Let be the supremum of S. Then, for all x
we have u < v. There exists a 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 is bounded. Show
that sup () = max {sup A, sup B}.
Solution: Let be real numbers such that and
| for all Define M = max {}. Then, for all
. So, is bounded. Further: Let sup A = u and sup B = v and
w = max{u, v}. For
Then, w is an upper bound of If z is an upper bound of then z
is an upper bound of A and B. And,
Therefore, w = ().

10. Let S be a non empty subset of IR that is bounded above. For any
define Prove that sup (a + S) = a+ sup S.
Solution: Let u = sup S, then for all . Then,
Thus, a + u is an upper bound for the set a + S. Thus,
Now, if v is any upper bound of the set a + S then for all
Consequently, for all so that v − a is an upper bound of S.
Then u = sup and so, Since v is any upper bound of
a + S and sup(a + S) is also an upper bound, we have

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