Page 38 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Axiom of Completeness of R, the subset S must have the supremum in R. We claim that this
supremum does not belong to Q.

Suppose m is the supremum of the set S. If possible, let m belong to Q. Obviously, then m > 0.
2
2
2
Now either m < 2 or m = 2 or m > 2.
2
Case (i) When m < 2. Then a number y defined as
+
4 3 m
y =
3 2 m
+
is a positive rational number and
2
2(2 m )
-
y – m =
+
3 2 m
2
2
Since m < 2, therefore 2 – m > 0, Hence
2
3(2 m )
-
y – m = > 0
+
3 2 m
which implies that y > m.
Again,
2
4 3 mö
æ +
y – 2 = ç ÷ - 2
2
+
è 3 2 mø
2
m - 2
= 2
(3 + 2 m)
Since m < 2, therefore
2
2
2
y – 2 < 0 i.e. y < 2.
This shows that y  S and also it is greater than m (the supremum of S). This is absurd. Thus the
case m < 2 is not possible.
2
Case (ii) When m = 2.
2
This means there exists a rational number whose square is equal to 2 which is again not possible.
Case (iii) When m > 2
2
In this case consider the positive rational number y defined in case (i). Accordingly, we have
2
2(2 m )
-
y – m = < 0 (check yourself)
3 2 m
+
i.e. y < m.
æ + 2 m 2
4 3 m)ö
-
Also 2 – y = 2 – ç ÷ 2 = 2
2
+
è 3 2 m ø (3 2 m)
+
2
i.e. 2 – y < 0 or y > 2,
2
which shows that y is an upper bound of S.

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