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  1. #1
    Founder Alavatar's Avatar
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    Smile Math proof. Find the error.

    Ok, so I like math. I found this simple proof when I was in college and thought it was nifty. But, there is an error in it that nullifies it's argument. Can you find it?

    a = b

    a^2 = a*b

    a^2 - b^2 = a*b - b^2

    (a + b)*(a - b) = b*(a - b)

    (a + b) = b

    b + b = b

    2b = b

    2 = 1

    If you have any other fun math problems or proofs list them! I realize that this probably isn't going to be a popular subject, but as a math geek I was hoping that perhaps others might think it is interesting.

  2. #2
    Community Member Darkschneider's Avatar
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    Quote Originally Posted by Alavatar View Post
    a = b
    <snip out the worthless steps>
    (a + b)*(a - b) = b*(a - b)
    (a + b) = b
    If a=b, then a - b = 0

    Can't divide by 0
    Quote Originally Posted by Shrike
    That mob was so like "I'm going to kill you" and you were like "Die".

  3. #3
    Community Member Darkschneider's Avatar
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    Step 1: -1/1 = 1/-1

    Step 2: Taking the square root of both sides:

    Step 3: Simplifying:

    Step 4: In other words, i/1 = 1/i.

    Step 5: Therefore, i / 2 = 1 / (2i),

    Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),

    Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

    Step 8: ,

    Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

    Step 10: and this shows that 1=2.

    Roll with the complex number fun.
    Quote Originally Posted by Shrike
    That mob was so like "I'm going to kill you" and you were like "Die".

  4. #4
    Hall of Famer Cupcake's Avatar
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    grumbles. I hate math. Pick a different subject LOL.
    Hands you a Cupcake One of Many of the O'Rum Ferretus's

  5. #5
    Founder Alavatar's Avatar
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    Quote Originally Posted by Darkschneider View Post
    Step 1: -1/1 = 1/-1

    Step 2: Taking the square root of both sides:

    Step 3: Simplifying:

    Step 4: In other words, i/1 = 1/i.

    Step 5: Therefore, i / 2 = 1 / (2i),

    Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),

    Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

    Step 8: ,

    Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

    Step 10: and this shows that 1=2.

    Roll with the complex number fun.
    Wow this one racked my brain a bit!

    I think the fault is in the proof for i/1 = 1/i?

    sqrt(1/-1) = sqrt(-1/1) -> i = i

    I am not completely positive, just confused. Because if you multiply (1/i)*(i/i) you get (-i), which is not i/1. Or perhaps complex numbers just throw a wrench in basic algebra?
    Last edited by Alavatar; 03-29-2007 at 07:01 PM.

  6. #6
    Founder Alavatar's Avatar
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    Quote Originally Posted by Darkschneider View Post
    If a=b, then a - b = 0

    Can't divide by 0
    Oh, and correct.

    Also, a^2 - b^2 = a*b - b^2 is 0 = 0. Factor 0 and you still are at 0. Divide 0/0 and you get infinity.

    Simple algebra, but it's fun.

    Hairy math = advanced electrical engineering stuff. Whoa, that stuff is beastly.
    Last edited by Alavatar; 03-29-2007 at 07:10 PM.

  7. #7
    Community Member DKerrigan's Avatar
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    Quote Originally Posted by Alavatar View Post
    Whoa, that stuff is beastly.
    But fun, if only I'd gotten here sooner...got anything else?
    Mr. Potatohead...MR. POTATOHEAD!!! Backdoors are NOT SECRETS!!!

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  8. #8
    Founder Alavatar's Avatar
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    Quote Originally Posted by DKerrigan View Post
    But fun, if only I'd gotten here sooner...got anything else?
    I don't. The one I posted is the only trick proof I have and I was hoping others might have stuff.

  9. #9
    Community Member ehondajim's Avatar
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    I feel..........not smart.
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  10. #10
    Founder joker965's Avatar
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    Quote Originally Posted by Alavatar View Post
    Ok, so I like math. I found this simple proof when I was in college and thought it was nifty. But, there is an error in it that nullifies it's argument. Can you find it?

    a = b

    a^2 = a*b

    a^2 - b^2 = a*b - b^2

    (a + b)*(a - b) = b*(a - b)

    (a + b) = b

    b + b = b

    2b = b

    2 = 1

    If you have any other fun math problems or proofs list them! I realize that this probably isn't going to be a popular subject, but as a math geek I was hoping that perhaps others might think it is interesting.
    Well I saw the accepted answer and I'm not really sure.

    My thinking is this.

    If (a + b) = b then a and b must be zero.

    If b+b=b then a and b must be zero.

    If 2b=b then a and b must be zero.

    2 can not equal 1

    Trying to solve a proof that can not possibly be correct will obviously lead to impossible operations.

    a = b

    a^2 = a*b

    a^2 - b^2 = a*b - b^2

    (a + b)*(a - b) = b*(a - b)

    0=0
    Anything that doesn't kill us can still hurt really bad.

    The Joker

  11. #11
    Founder Alavatar's Avatar
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    Quote Originally Posted by joker965 View Post

    <SNIP>
    Kind of. The problem with the proof is in the 3rd step.

    If a = b then a^2 - b^2 = 0 and ab - b^2 = 0, so anything after that is just manipulating the number 0.

  12. #12
    Community Member Yvonne_Blacksword's Avatar
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    Quote Originally Posted by Alavatar View Post

    a = b

    a^2 = a*b

    a^2 - b^2 = a*b - b^2
    if a=b then a^2=b^2 subtracting made left side =0
    if a=b then a*b=b^2 subtracting them made right side =0
    you made both sides =0...which is true...but sad.
    (a + b)*(a - b) = b*(a - b)
    a=b so subtracting a-b and multiplying through made left side =0 still ok...
    a=b so subtracting a-b and multiplying through made right side =0 ok..ok...
    (a + b) = b
    since a-b=0...you divided both sides by zero...you! you! you! You cant do that....
    you cant divide anything by zero and get any real answer...you are mean....I hate you!
    b + b = b
    BLllllTTTTttT
    2b = b
    wrong!
    2 = 1
    WrOngE SuX0R
    i hate this...
    Last edited by Yvonne Blacksword; 07-31-2007 at 07:26 PM.
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  13. #13
    Founder Alavatar's Avatar
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    Here is a practice question I came up with about 2 years ago (I had to resolve it to remember how to do it. ). I originally made it for my co-workers, none of them answered it except to say they had more important work to do!


    An airplane is cruising at an altitude of 2500ft and velocity of 500 ft/s. The airplane begins ascending along the parabolic path y = 0.0005x^2 + 2500 while it's air-speed indicator maintains the velocity of 500 ft/s. After traveling 1500 horizontal (x = 1500) feet the airplane releases it's cargo so now the cargo is in freefall. Assume no air resistance.

    A) If y=0 and the ground is assumed to be completely flat where does the cargo land?
    B) How long was the cargo in freefall?
    C) What is it's impact velocity in ft/s?
    D) What angle (in degrees) was the cargo released?
    E) What angle did the cargo impact the ground?


  14. #14

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    Quote Originally Posted by Alavatar View Post
    Here is a practice question I came up with about 2 years ago (I had to resolve it to remember how to do it. ). I originally made it for my co-workers, none of them answered it except to say they had more important work to do!



    Hmm, Just a question. If air resistance = 0 wouldn't flight be impossible?

  15. #15
    Founder Alavatar's Avatar
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    Quote Originally Posted by Vardak View Post
    Hmm, Just a question. If air resistance = 0 wouldn't flight be impossible?
    Technically, yes. But this is a math problem designed to facilitate the use of basic physics equations. Air resistance is considered a bit more advanced then 'basic' physics. Besides, I just didn't want to deal with ODEs.

    Edit: Also, with air resistance you need to specify a whole bunch of new data such as shape of the cargo, mass, rotation of the object in freefall (unless cargo is spherical), etc. A big PITA that I haven't had to deal with for about 5 years, now, and I don't really want to search through my books to relearn this right now.
    Last edited by Alavatar; 08-01-2007 at 07:51 PM.

  16. #16
    Community Member Tous's Avatar
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    I'm going to the dragonlance movie website and wash this math stuff from my brain.

  17. #17
    Community Member Reisei's Avatar
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    i didnt understand the responses that came after but this is what i thought when i first read it

    a = b/////a=a correct

    a^2 = a*b/////2a = 2a correct

    a^2 - b^2 = a*b - b^2/////2a - 2a = 0/////2a - 2a = 0 correct

    (a + b)*(a - b) = b*(a - b)///// 2a * 0 = 0/////a * 0 = 0

    (a + b) = b///// 2a = a not correct

    b + b = b///// not correct for the same reason

    2b = b///// not correct for the same reason

    2 = 1 im confused i dont think i know how proofs work i guess it doesnt help that ive been up all night

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