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Smug/flavorful proof phrases

  1. Jan 5, 2015 #1
    So, as all of you know, it is common in mathematical proof to begin a statement within the proof with one of those phrases such as "then," or "therefore," or "and so," or "hence", "thus" etc.

    But sometimes, for flavor, they can get a little more smug, such as,

    "indeed," - my topology professor. You had to hear him say it.
    "clearly," - pretty smug
    "it is clear that"

    then, we start getting into the more epic:

    "Bear witness to the fact that,"

    And these can even be combined:

    "Indeed, let us bear witness to the fact that"

    But perhaps the most epic one of all, was an algebra professor at my school (Arturo Magidin, for the math.SE posters):

    "We let x stand sentinel to the fact that"

    Wow, straight out of JRR Tolkein.

    So, for fun, I want to see what you can come up with. For the sake of participation, I have included a small proof in which you can "fill in the blanks" with the most flamboyant, pompous proof phrases you can think of.

    example:
    Let k be an even integer.
    Indubitably, there is exist an integer n such that k = 2n.
    To further embark on our quest for truth, let a be an integer. Then consider ak.
    It is abundantly clear as the full moon on a pale October night, ak = 2(an)
    Rightfully so, it is indeed shown for all to see that ak is even by definition
     
  2. jcsd
  3. Jan 5, 2015 #2

    Bystander

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    "_________" is trivially obvious and will be left as a exercise for the reader.
     
  4. Jan 6, 2015 #3
    Claim: Any multiple of an even integer is even.

    Let k be an even integer.
    Since k is stipulated to be even, it is patently obvious there must exist some integer, n, which conforms perfectly to the statement, k = 2n.
    Steamrolling logically and inexorably forward, let a be an integer. Then consider ak.
    It will be immediately apprehended, and is not rationally possible to doubt, that ak = 2(an)
    Therefore, we may boldly take up our chisel and hammer and vigorously inscribe in stone that ak is even by definition, and that, by extension to the infinite series of integers to which a, k, and n, belong, any multiple of an even integer is even, now, here, and in the whole universe, for all eternity.
     
  5. Jan 6, 2015 #4

    Bandersnatch

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    And lo, the Lord spake of k, and He named k an integer.
    Forsooth, there does exist an integer n. n is an integer, and by divine grace it doth satisfy k=2n.
    Then the Lord said: "Behold, I give a unto you. a I give you, and a is an integer. Venerate the holy union of ak."
    Thus spake the Lord. And great light descended upon the chosen people, and hosts of angels sang in unison, and they all saw with immaculate clarity that ak=2(an).
    And they spread their seed across the four winds and preached the Truth of ak being even. Such is the definition from on high.
     
  6. Jan 14, 2015 #5

    epenguin

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    It can be shown

    :headbang:
     
  7. Jan 16, 2015 #6
    In the beginning we have k as an even integer.
    We postulate that, there is another integer n such that k = 2n.
    Furthermore it can be shown that if we introduce a as an integer. We can obviously consider ak.
    And by all operators definitions ak would then = 2(an).
    Therefore and from thence on, ak will be even by definition QED
     
  8. Jan 19, 2015 #7
    Claim: Any multiple of an even integer is even.

    Proof:
    Let k be an even integer.
    As even the most vacuous student will immediately realize, there is quite obviously an integer n such that k = 2n.
    The next step is ridiculously elementary, and we omit it for brevity.
    Forsooth, it is immediately and irrevocably clear that beyond all doubt and question, ak = 2(an).
    The remainder of the proof is left as an incredibly trivial exercise.
     
  9. Jan 19, 2015 #8

    lisab

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    "It is left as an exercise for the reader..."

    I try to work this one into my daily life :D.
     
  10. Jan 19, 2015 #9
    As it is written, so let it be shown.
     
  11. Jan 19, 2015 #10
    Let K be an integer.
    It can be shown that K [has some property]
    It is now clear that [some property is related to some other property]
    We omit several elementary steps for clarity. (Yes, clarity).
    We leave the remainder of the proof as an exercise for the reader.
    QED.
     
  12. Jan 19, 2015 #11

    jbunniii

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  13. Jan 20, 2015 #12
    Check out my slightly more serious thread on the topic: https://www.physicsforums.com/threads/these-words-should-never-be-uttered-by-mathematicians.783296/

    lol
     
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