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kevo Group Title

Mathematical Proofs

  • 2 years ago
  • 2 years ago

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  1. mr.luna Group Title
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    nice

    • 2 years ago
  2. kevo Group Title
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    Prove that \[\exists z \in \mathbb{R} \forall x \in \mathbb{R}^+[\exists y \in \mathbb{R}(y-x+y/x) <--> x \neq z)\]

    • 2 years ago
  3. kevo Group Title
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    Lol, sorry for the wait, it was a pain to type up.

    • 2 years ago
  4. Wheaton71 Group Title
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    What is this

    • 2 years ago
  5. kevo Group Title
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    Proofs.

    • 2 years ago
  6. Wheaton71 Group Title
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    Teach me

    • 2 years ago
  7. eliassaab Group Title
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    What does <−−> mean?

    • 2 years ago
  8. kevo Group Title
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    its the symbol for if and only if.

    • 2 years ago
  9. eliassaab Group Title
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    It is still not clear to me what the last statement mean. \[ \exists z \in \mathbb{R} \forall x \in \mathbb{R}^+[\exists y \in \mathbb{R} \] such that what?

    • 2 years ago
  10. Wheaton71 Group Title
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    Yeah I'd also like to know.

    • 2 years ago
  11. kevo Group Title
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    What it says after R?

    • 2 years ago
  12. KingGeorge Group Title
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    just fyi, if you want to increase the space between characters in the equation editor, simply type "\;" for a small space, "\:" for a slightly bigger one, "\quad" for a big one, and "\qquad" for a giant one. This helps increase readability.

    • 2 years ago
  13. kevo Group Title
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    Oh, I had no idea!

    • 2 years ago
  14. kevo Group Title
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    Thanks for the tip :)

    • 2 years ago
  15. eliassaab Group Title
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    Yes after R

    • 2 years ago
  16. KingGeorge Group Title
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    I think the best way to approach this would be to split it into two parts. First we want to show implication tot he right, and second we want to show implication to the right. Also note that \[p \Rightarrow q \quad \Longleftrightarrow \neg \;p \;\;\text{V}\;\;q\]

    • 2 years ago
  17. KingGeorge Group Title
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    So to show implication to the right, let's see if we can prove the simpler statement.

    • 2 years ago
  18. KingGeorge Group Title
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    I'm a little confused about the statement \((y-x+y/x)\). What is it saying? In this form it's virtually meaningless.

    • 2 years ago
  19. kevo Group Title
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    That's a good question.. so there exists y in R(y - x = y/x) iff x does not equal z.

    • 2 years ago
  20. kevo Group Title
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    This is just one confusing statement that should not be legal to give to student. Just saying.

    • 2 years ago
  21. KingGeorge Group Title
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    There's supposed to be an equals sign there. That helps. Give me a second to think about this.

    • 2 years ago
  22. kevo Group Title
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    LOL i just realized that I mistyped that. Sorry!

    • 2 years ago
  23. KingGeorge Group Title
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    Let's show implication to the right first. To show this, we need to choose a z such that for all y and x (x positive) \((y-x =y/x)\) or \(x\neq z\). Just choose z to be negative. Since x is positive, we know that \(x \neq z\).

    • 2 years ago
  24. KingGeorge Group Title
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    Now we need to show implication to the left. To show this, we need to choose a z such that for all positive x, there exists a y such that that \(z=x\) or that \(y-x\neq y/x\) Here, just choose \(y=0\). Since x is positive, \(0-x\) is less than 0, and \(0/x=0\).

    • 2 years ago
  25. KingGeorge Group Title
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    Therefore, we are done. Sorry that took a while for me to write. Also, Instead of writing "we need to choose a z such that for all y and x (x positive)" in the first part, I should have written "we need to choose a z such that for all positive x there exists a y" It doesn't really matter in the end however.

    • 2 years ago
  26. kevo Group Title
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    KingGeorge. You are amazing. You are seriously my hero. I don't know how you are so good at this, but thank you.

    • 2 years ago
  27. KingGeorge Group Title
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    Practice, and a little bit of natural skill is how I'm good. I've also had some amazing teachers.

    • 2 years ago
  28. kevo Group Title
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    I know who I'm asking for help on proofs from now on :). I just don't know how I can reward you..

    • 2 years ago
  29. KingGeorge Group Title
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    As long as you're trying to learn, I'll be good.

    • 2 years ago
  30. kevo Group Title
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    Well, if you're ever in Seattle, I'll buy you dinner.

    • 2 years ago
  31. KingGeorge Group Title
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    Sounds good. :)

    • 2 years ago
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