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anonymous

  • 5 years ago

Someone who is good with proofs please talk to me

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  1. anonymous
    • 5 years ago
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    I am not good, but I will try the best I can.

  2. anonymous
    • 5 years ago
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    ok I will upload the problem

  3. anonymous
    • 5 years ago
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  4. anonymous
    • 5 years ago
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    which problem?

  5. anonymous
    • 5 years ago
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    1, 2, and 3 they are extra credit

  6. anonymous
    • 5 years ago
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    Couldn't you just pick an a=-9 and see if square root of -9^2 is the same as |-9| ?

  7. anonymous
    • 5 years ago
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    R means any number, so try to pick numbers that you think would make it bad, and if they are good..use that as proof.

  8. anonymous
    • 5 years ago
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    But if you are looking for book definition type proofs you got me.

  9. anonymous
    • 5 years ago
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    I think that when it says show that it wants you to pick numbers from the real set and try to make it false. Is that your understanding too?

  10. anonymous
    • 5 years ago
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    well.. I think I got the solution of the first problem

  11. anonymous
    • 5 years ago
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    if it involves differentail equations, im outta here :)

  12. anonymous
    • 5 years ago
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    i thought your suppose to prove what he asks for

  13. anonymous
    • 5 years ago
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    proof is a relative term.

  14. anonymous
    • 5 years ago
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    proof to a college teacher and proof to a 6 year old are not going to be the same.

  15. anonymous
    • 5 years ago
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    I think anny is doing a differential equation up there :)

  16. anonymous
    • 5 years ago
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    since a^2=(+a)^2=(-a)^2, the numbers a, -a are square roots of a^2.. that's: \[a \ge0 \implies \sqrt{ a^2}=a\] \[a<0 \implies \sqrt {a^2}=-a\] which is the same as the definition of the absolute value, so \[\sqrt {a^2}=\left| a \right|\]

  17. anonymous
    • 5 years ago
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    you know the definition of the absolute value. right?.. does that make sense then?

  18. anonymous
    • 5 years ago
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    we can use what's this theorem we just proved to prove the next problem.

  19. anonymous
    • 5 years ago
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    oh ok

  20. anonymous
    • 5 years ago
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    we want to prove that: \[\left| ab \right|=\left| a \right|.\left| b \right|\] but we just proved that: \[\left| ab \right|=\sqrt {(ab)^2}\] we can now proceed, \[\sqrt {(ab)^2}=\sqrt {a^2b^2}=\sqrt {a^2} . \sqrt{b^2}=\left| a \right|.\left| b \right|\] that's it.

  21. anonymous
    • 5 years ago
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    I dunno if that makes any sense to you, but that's what I have anyway. I wish you get some points for that :)

  22. anonymous
    • 5 years ago
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    where did you get the \[\sqrt{ab ^2{?}}\]

  23. anonymous
    • 5 years ago
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    yes, me too thank you

  24. anonymous
    • 5 years ago
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    it's the first theorem we proved, but instead of a, we have ab.

  25. anonymous
    • 5 years ago
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    but how does it become ab^2 when its just IabI= IaI IbI

  26. anonymous
    • 5 years ago
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    From part one you showed that \[|x| = \sqrt{x^2}\] Let x = ab and you have it.

  27. anonymous
    • 5 years ago
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    what about # 3

  28. anonymous
    • 5 years ago
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    the third one is easy.. I will give you a hint

  29. anonymous
    • 5 years ago
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    good I wanna try

  30. anonymous
    • 5 years ago
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    the following is known: \[a \le \left| a \right|, b \le \left| b \right|\]

  31. anonymous
    • 5 years ago
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    Just have to go through the combinations\(a\ge 0, a \lt 0, b \ge 0, b \lt 0\) I'm not sure you can assume that \(x \le |x|\) otherwise you'd just let x = ab and you've already got it. I think you have to show that \(x \le |x|\) using the definition of the absolute value.

  32. anonymous
    • 5 years ago
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    ^^ I totally agree

  33. anonymous
    • 5 years ago
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    though it's easy to prove that using the definition of the absolute value.

  34. anonymous
    • 5 years ago
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    what topic is this, thats how lost I am in my calculus class :(

  35. anonymous
    • 5 years ago
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    Ok, if a \(\ge\)0 then what is |a| ?

  36. anonymous
    • 5 years ago
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    hey lisaj you have to go back and see the definition of the absolute value.

  37. anonymous
    • 5 years ago
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    all proofs will make no sense to you if you are not aware of the definition.

  38. anonymous
    • 5 years ago
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    @ polpak: IaI is idk i dont get it

  39. anonymous
    • 5 years ago
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    if a is 5 what is lal?

  40. anonymous
    • 5 years ago
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    the absolute value is 5

  41. anonymous
    • 5 years ago
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    good. what if a=-5, then its absolute value is?

  42. anonymous
    • 5 years ago
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    5

  43. anonymous
    • 5 years ago
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    So if a is less than 0, |a| = -a. If a is greater than (or equal to) 0, |a| = a

  44. anonymous
    • 5 years ago
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    Right?

  45. anonymous
    • 5 years ago
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    true

  46. anonymous
    • 5 years ago
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    so you can see now that |a| is always greater than or equal to a?

  47. anonymous
    • 5 years ago
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    so # 3 would be : ab\[ab \le \ \-a |-b\|]

  48. anonymous
    • 5 years ago
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    yes anny

  49. anonymous
    • 5 years ago
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    ab < or equal to I-aII-bI

  50. anonymous
    • 5 years ago
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    \[x \ge 0 \implies |x| =x \implies |x|\ge x\] \[x \lt 0 \implies |x| = -x \implies |x| \ge x\] \[\implies |x| \ge x \text{,}\forall x \in R\] Let ab be a product of real numbers \[\implies ab \in R \implies |ab| \ge ab \implies ?\]

  51. anonymous
    • 5 years ago
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    \[\left| ab \right|\ \ge \left| -a \right|\left| -b \right|\]

  52. anonymous
    • 5 years ago
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    Well sure, but |a| = |-a|

  53. anonymous
    • 5 years ago
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    is what i did the correct answer?

  54. anonymous
    • 5 years ago
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    You are trying to show that \(ab\le |a||b|\)

  55. anonymous
    • 5 years ago
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    so then it would be |−a||−b|≤ab

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