But if |y-y_0|<\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )
then |y-y_0|< \frac {\epsilon}{2|x_0|+1}
Since if \textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )=\frac {\epsilon}{2|x_0|+1} then |y-y_0|<\frac {\epsilon}{2|x_0|+1}\leq 1.
And if \textrm{min} \bigg ( \frac...
Homework Statement
Prove that if |x-x_0| < \textrm{min} \bigg ( \frac{\epsilon}{2|y_0|+1},1 \bigg ) and |y-y_0| < \frac{\epsilon}{2|x_0|+1} then xy-x_0y_0<\epsilon
Homework Equations
We can use basic algebra and the following axioms:
For any number a, one and only one of the following...
Is there more than one meaning of the notation "f(x)=[x]"?
In my real analysis textbook there is a question that says:
Decide whether f(x)=[x] is bounded above or below on the interval [0,a] where a is arbitrary, and whether the function takes on it's maximum or minimum value within that...
Figured it out!
(1) Since b \in P and (d - c) \in P, we get (bd - bc) \in P using P12.
(2) Since c \in P and (b - a) \in P, we get (bc - ac) \in P using P12.
Then from (1) and (2) we get (bd - bc + bc - ac) \in P using P11.
Using P3 and P2 we get (bd - ac) \in P, thus...
Homework Statement
Prove that for all numbers a, b, c, d: if 0 \leq a < b and 0 \leq c < d then ac < bd.
This is problem 5 from chapter 1 of Michael Spivak's "Calculus", 4th Edition. It is the text for my real analysis course.
I should also mention that this is not a homework problem...