Math Proof

Let x and y be real numbers. Then x=y if and only if abs(x-y)< epsilon for every epsilon>0

  1. 👍 0
  2. 👎 0
  3. 👁 159
asked by Ashley
  1. LOL, well I suppose one could prove it somehow. It seems to belabor the obvious though.

    1. 👍 0
    2. 👎 0
    posted by Damon
  2. I honestly have no clue how to even start this proof. Other than assume x=y and then assume that abs(x-y)<epsilon

    1. 👍 0
    2. 👎 0
    posted by Ashley
  3. well, draw two points on a number line maybe a distance epsilon apart then let epsilon go to zero.

    1. 👍 0
    2. 👎 0
    posted by Damon
  4. on second thought draw x as a point on a number line, then y points epsilon to the right and to the left. Then let epsilon ---> 0

    1. 👍 0
    2. 👎 0
    posted by Damon
  5. I'm still not seeing it. I have not used epsilon before today

    1. 👍 0
    2. 👎 0
    posted by Ashley
  6. I know that if x=y, for any epsilon>0, abs(x-y)=0<epsilon

    1. 👍 0
    2. 👎 0
    posted by Ashley
  7. well epsilon is anything you want it to be, but usually you make it a small number and let it approach zero.
    If x = 5
    and y = 7
    then |x-y| = epsilon
    now we would call epsilon = 2
    BUT
    x is only equal to y if |x-y|< ANY epsilon bigger than zero, including e = .0000000001
    so when could |x-y| be that small?
    Only if x is awfully close to y
    but epsilon could be much smaller than .0000000000001, in fact our equality is only true if epsilon is so small we can not tell it is there at all.

    1. 👍 0
    2. 👎 0
    posted by Damon
  8. Okay, so since epsilon has to be the smallest number passible x has to equal y.

    1. 👍 0
    2. 👎 0
    posted by Ashley
  9. yes
    it has to be for any old epsilon, including one that is vanishingly small so that the difference between x and y is so tiny it does not exist.

    1. 👍 0
    2. 👎 0
    posted by Damon
  10. How would I write that though, since epsilon has to be greater than 0 and abs(x-y) has to be less than epsilon. would i suppose that x doesn't equal y, then epsilon <0.. then epsilon is less than abs(x-y)/2, then abs(x-y)<epsilon<abs(x-y)/2. which contradicts abs(x-y)< epsilon for all epsilon<0.

    1. 👍 0
    2. 👎 0
    posted by Ashley
  11. You say that as epsilon ---> 0, |x-y|--->0

    1. 👍 0
    2. 👎 0
    posted by Damon
  12. Thank you

    1. 👍 0
    2. 👎 0
    posted by Ashley
  13. The key word here is ** EVERY ** epsilon, including one vanishingly small .

    1. 👍 0
    2. 👎 0
    posted by Damon
  14. You are welcome.

    1. 👍 0
    2. 👎 0
    posted by Damon

Respond to this Question

First Name

Your Response

Similar Questions

  1. C programming, not C++

    design and implement such a function and use it in a program called part3.c that allows someone to enter two real numbers and an epsilon value (also a real number) and reports if the two numbers are equal (+/- epsilon) or not. If

    asked by apoorva on April 16, 2008
  2. how to understand this

    design and implement such a function and use it in a program called part3.c that allows someone to enter two real numbers and an epsilon value (also a real number) and reports if the two numbers are equal (+/- epsilon) or not. If

    asked by apoorva on April 16, 2008
  3. Math

    let x and y be real numbers. find the area enclosed by the graph of the following equation: abs(2x)+abs(2y)=1

    asked by Zee on April 13, 2011
  4. math

    let x and y be real numbers. find the area enclosed by the graph of the following equation: abs(2x)+abs(2y)=1

    asked by Diana on April 13, 2011
  5. real analysis

    i noticed it doesn't have x in them. i don't know what to do. prove: let epsilon>0. let f(x) = pi + 3 for x∈ real number. Prove your conjecture. Whether f is continuous for all r ∈ real number. my proof;

    asked by Lucy on September 20, 2014
  6. Math

    Use a graph to find a number delta such that if abs(x-pi/4) < delta then abs(tanx-1) < 0.2. I understand that the epsilon is 0.2. But what step do I take next? How do I solve the problem?

    asked by KC on September 1, 2012
  7. Math Proof

    Prove that for all integers n > or = 1 and all real numbers x subscript 1, x subscript 2,...., x subscript n. absolute value of (x subscript 1+ x subscript 2+...+ x subscript n)

    asked by Ashley on September 1, 2014
  8. math

    would this be a correct proof of the cauchy-schwartz inequality: abs=absolute value abs(u*v) is less than or equal to abs(u)*abs(v). Then you divide both sides by abs(u)*abs(v) so that you get cos(theta) is less than or equal to

    asked by annie on May 14, 2009
  9. Int Algebra

    The difference of two numbers is 7. The sum of the numbers is -1. What are the numbers? I've come up with the following equations: x-y=7 and x+y=-1. I can't seem to solve because there are no numbers on the left side of the

    asked by Rose Mata on January 26, 2007
  10. Geometry

    Show the conhecture is false by finding a counterexample... 1. The differnece of the absolute value of two numbers is positive meaning /a/ - /b/ > 0 I do not get that.... =/ HELP. If absolute value of a is greater than the

    asked by Katy on August 14, 2007

More Similar Questions