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I Check my logic

  1. Jan 22, 2017 #1
    Given ##a,b\inℝ## with ##a≥b## and ##b≥c##, I wish to show that ##a≥c##. For that we have ##a-b≥0## and ##b-c≥0##. Therefore, ##a-b+b-c≥0## and so ##a-c≥0##. Hence, ##a≥c##. Q.E.D.

    Correct me if I am wrong, but this works unless I should use different cases because of equality and inequality. What do you think?
     
  2. jcsd
  3. Jan 22, 2017 #2
    Looks good to me. Even simpler, this can be written as a≥b≥c, and automatically, a≥c.
     
  4. Jan 22, 2017 #3

    micromass

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    How does this prove anything. You just restate the result which needs to be proven.
     
  5. Jan 23, 2017 #4

    mfb

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    How do you get from ##a \geq b## to ##a-b \geq 0## and from (##a-b \geq 0## and ##b-c \geq 0##) to ##a-b +b-c \geq 0##?

    At that level, you have to do everything with the definitions and axioms.
     
  6. Jan 23, 2017 #5
    It shows that if a is greater than/equal to b, then it must be greater than/equal to whatever b is greater than/equal to: in this case, c.
     
  7. Jan 23, 2017 #6

    micromass

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    You're assuming what requires to be shown: transitivity.
     
  8. Jan 23, 2017 #7
    That being said, how would you prove it?
     
  9. Jan 23, 2017 #8

    micromass

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    Depends on the definition and the axioms given to me.
     
  10. Jan 27, 2017 #9
    My proof as written is incomplete. In the final proof I'd be sure to justify my assertions using definitions and axioms.
     
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