Real Analysis Answers

show if a,b ЄR then,
a. max (a,b) = (1/2)(a+b+|a-b|) and min (a,b)=(1/2)(a+b-|a-b|)
b. min (a,b,c)=min(min(a,b),c)

Find all x ∈ R that satisfy the equation& |x+1| + |x-2| = 7.

If& x, y, z& ∈ R and& x ≤ z, show that& x ≤ y ≤ z& if & |x-y| + |y-z| = |x-z|.& & Interpret this geometrically.

Let a,b є R, and suppose that for every ε > 0 we have a ≤ b+ε. show that a ≤ b.

Show that if& a < b, then a< (1/2)(a+b) < b&

Show that if a > 0, then (1/a) > 0 and (1/(1/a)) = a.

Show that there does not exist a rational number t such t[sup]2[/sup] = 3.

If a ≠ 0 and b ≠ 0, show that (1/ab) = (1/a)(1/b).

If 0 < a < b, show that
1)& a < (ab)[sup]1/2[/sup] < b;
2) 1/b < 1/a.

If a,b ϵ R, show that a[sup]2[/sup] + b[sup]2[/sup] = 0, if and only if a = 0 and b = 0