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I can't start proofs

  1. Feb 21, 2006 #1
    everything i have to prove seems impossible
    then i see it done and it seems so easy
    any help on starting a proof,...anybody ever have this problem
    i am new to this stuff, but my problem is I don't know what i need to show and what is legal to use
    so this is my current problem, i'm sure it's extremely easy for all you math geniuses so spare me the arrogance

    If f itself is 1-to-1 then f^-1 is a function.
    Let f:G->H be an onto function
    (i) If V is contained in H then f(f^-1(V))=V

    i'm pretty sure i understand onto and 1-1 definition
    i also understand why it's true, i think
    i also can somewhat understand the homomorphism stuff of it

    i just don't know where to start, what to do, what it really is asking

    then i gotta prove this, but have not looked into it because i'm still on (i)
    (ii) If W is contained in G then W is contained in f^-1(f(W)) and this could be a strict containment

    abstract algebra seems to be an endless barrier in my math knowledge
    is the key to it, memorizing endless amounts of definitions??:yuck:
  2. jcsd
  3. Feb 21, 2006 #2
    1) Write down the definitions. (that's how you start!)
  4. Feb 21, 2006 #3
    i did, i wrote evrerything down i knew
    but i'm not sure how to apply it i guess, (that's how i started!!)
  5. Feb 21, 2006 #4


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    anatomy of a proof

    As you see more and more proofs, you'll start to see that there is a finite number of tehcniques one uses to prove things.

    But almost all of them begin by writing down the definitions.

    Start with the definition of an onto function from G to H.

    Follow with the definition of f^-1(V).

    Then with the definition of f(A) where A in a subset of G.

    Then replace A by f^-1(V).

    Just by applying the definitions you'll have what f(f^-1(V)) is. And if it is indeed V, that's what you'll find.
  6. Feb 21, 2006 #5
    thanks, i think that may make sense breaking it all down like that
  7. Feb 21, 2006 #6


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    heres one suggestion: try to refrain from being defensive and rude to people you are asking for help. i was going to help but was put off by the "spare me the arrogance" comment.
  8. Feb 21, 2006 #7


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    Wow, I didn't even notice that until you mentionned it.

    Those are harsh words.
  9. Feb 22, 2006 #8

    matt grime

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    no, that is not true. only if f is a bijection (one to one and onto) is there an inverse function. Perhaps that sentence ought to have come after this one.

    the most useful way to prove set equality is to show each is contained in the other. Pick something in V, and shwo that it is in f(f^-1(V)), which is obvious since for v in V v=ff^-1(v) by the definition of inverse function: g is inverse to f if and only if fg and gf are the identity maps. The other containment is as easy to show. Arguably too easy in some sense and it genuinely isn't easy to write the proof in these cases: it isn't clear what constitutes a solid argument and what doesn't when the solid argument is very close to just saying 'well, it is'.

    where do homomorphisms come into anything here?

    if you understand the inverse image of a set you will see that obviously more than just W may map into f(W) (we're now droppping the hypotheses on f being injective I take it) and certainly W maps into f(W). Provide a counter example for the second part.

    you've memorized thousands of words; i suggest you didn't hat learning french at school because it required vocabulary. why should this be any different?
    as it is you only need to memorize very few things compared to the knowledge you'll be required to keep in your head for everything else in this life.
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