Recent content by objectivesea

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    Prove Inequality: |x-x_0|, |y-y_0| & xy-x_0y_0<\epsilon

    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...
  2. O

    Prove Inequality: |x-x_0|, |y-y_0| & xy-x_0y_0<\epsilon

    I just double checked. I'm sure.
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    Prove Inequality: |x-x_0|, |y-y_0| & xy-x_0y_0<\epsilon

    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...
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    Undergrad Is there more than one meaning of the notation f(x)=[x] ?

    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...
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    Can We Prove ac < bd Under Given Conditions?

    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...
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    Can We Prove ac < bd Under Given Conditions?

    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...