# 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, r^{2} = 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

4k^{2} = 3n^{2} 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,

2s^{2} + 2s = 6t^{2} + 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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