Page 4 of 6 FirstFirst 123456 LastLast
Results 46 to 60 of 86

Math Help - Trick Questions

  1. #46
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Quick
    The first compass originated in China. Although I'm not sure where it was first practically used.
    If they were not sure where they were when the compass was first used it
    could not have been working properly

    RonL
    Follow Math Help Forum on Facebook and Google+

  2. #47
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    The "compass" was actually a toy, a "magic spoon" if you will. It was a spoon with magnets on the head so that if you dropped it the head would always point in the same direction.
    Quote Originally Posted by CaptainBlack
    If they were not sure where they were when the compass was first used it
    could not have been working properly

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #48
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Wow, this is stupid. Here is another one of my riddles.

    If G is an abelian group then,
    (a+b)^n=a^n+b^n.

    But wait! \mathbb{Z} is an abelian group yet, (1+1)^2=1^2+1^2=2 how can that be?
    Follow Math Help Forum on Facebook and Google+

  4. #49
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Wow, this is stupid. Here is another one of my riddles.

    If G is an abelian group then,
    (a+b)^n=a^n+b^n.
    What is the group operation here, and what does a^n denote?

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #50
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CaptainBlack
    What is the group operation here, and what does a^n denote?

    RonL
    The accepted notion of a^n represents an element of a group operated upon itself n-times. If n=1 then the result is the element itself. If n=0 then the result is the identity element. If n<0 then the result is the inverse of result when without the negative.
    Follow Math Help Forum on Facebook and Google+

  6. #51
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    The accepted notion of a^n represents an element of a group operated upon itself n-times. If n=1 then the result is the element itself. If n=0 then the result is the identity element. If n<0 then the result is the inverse of result when without the negative.
    Then does + represent the group operation, and do exponentiation and
    addition still mean this when you write:

    <br />
(1+1)^2=1^2+1^2=2<br />

    ??

    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #52
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CaptainBlack
    Then does + represent the group operation,
    Of course + is the group operation, surly multiplication does not form a group.

    Quote Originally Posted by CaptainBlank
    and do exponentiation and
    addition still mean this when you write:

    <br />
(1+1)^2=1^2+1^2=2<br />

    ??

    RonL
    I cannot answer that because it is as if I am giving an answer.
    Follow Math Help Forum on Facebook and Google+

  8. #53
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    544
    Thanks
    12
    If G is an abelian group then...
    I don't think this is enough. We need G to be a field (so addition and multiplication are both defined) and we need it to be of characteristic n:
    This means, ng=g+...+g=0 for all g in G.

    Under these considerations, (a+b)^n=a^n+b^n+(remaining terms), and since the remaining terms are all multiples of n, we have that (a+b)^n=a^n+b^n.

    Examples of such fields are the Z_p for p prime, as nx=0 for all x in Z_p.
    Follow Math Help Forum on Facebook and Google+

  9. #54
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by Rebesques
    I don't think this is enough. We need G to be a field (so addition and multiplication are both defined) and we need it to be of characteristic n:
    This means, ng=g+...+g=0 for all g in G.

    Under these considerations, (a+b)^n=a^n+b^n+(remaining terms), and since the remaining terms are all multiples of n, we have that (a+b)^n=a^n+b^n.

    Examples of such fields are the Z_p for p prime, as nx=0 for all x in Z_p.
    I was just about to say that
    Follow Math Help Forum on Facebook and Google+

  10. #55
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Rebesques
    I don't think this is enough. We need G to be a field...
    Nope.
    (a+b)^n=(a+b)+...+(a+b)=(a+..+a)+(b+...+b) (because G is an abelian group). Thus, a^n+b^n.
    But, nobody answer my problem, why is it that (1+1)^2\not = 1^2+1^2???
    Follow Math Help Forum on Facebook and Google+

  11. #56
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Of course + is the group operation, surly multiplication does not form a group.


    I cannot answer that because it is as if I am giving an answer.
    These are directed questions to show you are confusing two different
    operations in the case of the field \mathbb{Z} with the one
    you have in the case of a group.

    RonL
    Follow Math Help Forum on Facebook and Google+

  12. #57
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CaptainBlack
    These are directed questions to show you are confusing two different
    operations in the case of the field \mathbb{Z} with the one
    you have in the case of a group.

    RonL
    Say better. Exactly why it that wrong?
    Follow Math Help Forum on Facebook and Google+

  13. #58
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Say better. Exactly why it that wrong?
    If G is an abelian group then,
    (a+b)^n=a^n+b^n.
    Here there is but one operation:

    x^n=x \oplus x^{n-1};\ x^1=x

    and your relation should be:

    (a \oplus b)^n=a^n \oplus b^n

    where \oplus is the group's operator.

    But wait! \mathbb{Z} is an abelian group yet, (1+1)^2=1^2+1^2=2 how can that be?
    but here there are two distinct operations involved +, \times,
    as here x^2=x \times x and not x+x.

    RonL
    Follow Math Help Forum on Facebook and Google+

  14. #59
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CaptainBlack
    Here there is but one operation:

    x^n=x \oplus x^{n-1};\ x^1=x

    and your relation should be:

    (a \oplus b)^n=a^n \oplus b^n

    where \oplus is the group's operator.



    but here there are two distinct operations involved +, \times,
    as here x^2=x \times x and not x+x.

    RonL
    Let me say it my way.
    In the case of \mathbb{Z} the meaning of,
    (a+b)^n is actually,
    n(a+b). Thus,
    n(a+b)=na+nb are nothing is violated.

    The purpose of my riddle was to confuse the exponent. The meaning of exponent in this case is operating an element by itself a certain number of times. However, I confused it with the meaning of regular exponents of multiplication.
    Follow Math Help Forum on Facebook and Google+

  15. #60
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Here I came up with another riddle, this one seems really easy.

    Let X be a non-empty set and let ~ be a relation on X. Why is it that we say need to check reflexisve property if we can show it is it true from symettry and transitivity? Observe, if x~y then y~x (symettric) but then, x~x (transitive). Therefore it is reflexsive!?!?
    Follow Math Help Forum on Facebook and Google+

Page 4 of 6 FirstFirst 123456 LastLast

Similar Math Help Forum Discussions

  1. More dice trick questions
    Posted in the Statistics Forum
    Replies: 10
    Last Post: February 8th 2011, 03:13 AM
  2. Trick 4 solving 4 n
    Posted in the Algebra Forum
    Replies: 9
    Last Post: November 8th 2010, 05:46 PM
  3. trick to solve?
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: June 18th 2010, 10:48 AM
  4. Simple if not fot the Trick!
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: February 22nd 2009, 06:00 AM
  5. Conjurer's trick
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 20th 2007, 10:41 AM

Search Tags


/mathhelpforum @mathhelpforum