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    (Original post by sherunsaway)
    Moron. You can't divide both by (b-a) because there isn't a (b-a) on each side. Look carefully on the right side and there is (a+b) and (a-b), neither of which is the same as (b-a). All this example proves is how good your eyesight is.
    (Original post by MR_JR)
    ^^
    read this part


    thus his proof is incorrect
    Here you go. both of you, i corrected my equation. I wrote it correctly on paper, but just typed it incorrectly here (seems more prone to mistyping maths on computer!!!). i apologise for the inconvenience.
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    ^^
    no worry. It wasnt a dig at you, but the person who started with "oh it is right"

    now its corrected, tis funkeh
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    (Original post by MR_JR)
    ^^
    no worry. It wasnt a dig at you, but the person who started with "oh it is right"

    now its corrected, tis funkeh
    cheers.

    [btw ppl wondering how it is possible - the divide by (b-a) is wrong, as if a=b, therefore b-a=0, and you can't divide by 0! but most people dont notice!]
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    (Original post by dvs)
    How about this one, it uses the imaginary number i^2=-1:
    -1 = -1
    1/-1 = -1/1
    sqrt(1/-1) = sqrt(-1/1)
    1/i = i/1
    1 = i^2
    1 = -1

    problem here "sqrt(1/-1) = sqrt(-1/1)"

    when you sqrt a negative fraction, the numerator (top bit) always becomes the negative. So square rooting 1/-1 or -1/1 always returns i/1.

    if your calculator can be put into complex mode, try putting in 1/i, it probably will give you an error.

    i think thats right.
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    (Original post by Zuber)
    cheers.

    [btw ppl wondering how it is possible - the divide by (b-a) is wrong, as if a=b, therefore b-a=0, and you can't divide by 0! but most people dont notice!]
    you can always divide by 0!
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    (Original post by phreek)
    problem here "sqrt(1/-1) = sqrt(-1/1)"

    when you sqrt a negative fraction, the numerator (top bit) always becomes the negative. So square rooting 1/-1 or -1/1 always returns i/1.

    if your calculator can be put into complex mode, try putting in 1/i, it probably will give you an error.

    i think thats right.
    no, it's just people forget that -1 has two squareroots, +/- i
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    (Original post by homoterror)
    no, it's just people forget that -1 has two squareroots, +/- i
    i thought about that, but that applies for √1 too, so you'd have both sides as + or -
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    (Original post by phreek)
    i thought about that, but that applies for root √1 too, so you'd have both sides as + or -
    yes that is obvious.

    \font{3}i=\pm\sqrt{-1},\quad 1=\pm\sqrt{1}
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    it's like b x 0 = a x 0 = 0 where u can replace a and b with whatever u like ....
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    here's one (which may have already been posted, i admit i haven't looked)

    1/4 > 1/8
    (1/2)² > (1/2)³
    log (1/2)² > log (1/2)³
    2 log(1/2) > 3 log(1/2)
    2 > 3

    QED :eek:

    lol, its easy enough to figure it out but its quite neat
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    (Original post by MC REN)
    here's one (which may have already been posted, i admit i haven't looked)

    1/4 > 1/8
    (1/2)² > (1/2)³
    log (1/2)² > log (1/2)³
    2 log(1/2) > 3 log(1/2)
    2 > 3

    QED :eek:

    lol, its easy enough to figure it out but its quite neat
    \log (\frac{1}{2}) = -\log 2 < 0
    Is it just my browser or do the TeX fonts look really awful on this forum?

    no, it's just people forget that -1 has two squareroots, +/- i
    Don't we define i=\sqrt{-1} ?
    \sqrt{-1}=i though z^{2}=-1 has the two solutions z=\pm i
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    (Original post by Gaz031)
    \log (\frac{1}{2}) = -\log 2 < 0
    Is it just my browser or do the TeX fonts look really awful on this forum?
    yer, well done
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    I really feel an urge to shoot whoever it was that revived this thread
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    a glass with a capacity of 1 litre has 0.5 litres in it - ppl often say it is either half empty of half full and that pessimists choose half empty and optimists choose half full

    'the glass is half empty' FALSE

    an empty glass has 0 litres inside

    half of empty is 0.5 x 0 = 0 which doesn't equal 0.5

    'the glass is half full' TRUE

    a full glass has 1 litre in it

    half full is 0.5 x 1 = 500mml

    therefore a 1 litre glass with 0.5litres in it is not half empty, but half full
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    What would appease you?

    Half Drained?
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    (Original post by esx77)
    a glass with a capacity of 1 litre has 0.5 litres in it - ppl often say it is either half empty of half full and that pessimists choose half empty and optimists choose half full

    'the glass is half empty' FALSE

    an empty glass has 0 litres inside

    half of empty is 0.5 x 0 = 0 which doesn't equal 0.5

    'the glass is half full' TRUE

    a full glass has 1 litre in it

    half full is 0.5 x 1 = 500mml

    therefore a 1 litre glass with 0.5litres in it is not half empty, but half full
    im one of them dumb ones i just realized. But it still makes sense doesnt it?
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    (Original post by MR_JR)
    its wrong. You cant divide.....

    DAMNIT sherunsaway you beat me to it!!!!


    Old classic


    x=0.9reccuring
    *10
    10x=9.9 reccuring
    10x-x
    9x=9
    x=1

    But x also equals 1, meaning that 1 ≡ 0.9reccuring


    Im thinking though, could this have the same principle as an ordered pair, in that to fit the quation there are 2 possible values. Actually maybe not, as they dont proooveeach other. Hmm i dunno
    dude. 1 is 0.9 recurring.
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    Wrong.

    However infinately small the differencee between 0.9rec and 1, there IS a difference. In this case it would be like

    0.01 with an infinite number of zeros
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    (Original post by MR_JR)
    Wrong.

    However infinately small the differencee between 0.9rec and 1, there IS a difference. In this case it would be like

    0.01 with an infinite number of zeros
    No there isn't.
    There's no such thing as 0.00....(rec).....1. You can't have any number after this infinite number of zeros.
    0.9rec and 1 are one and the same. (notice the unintentional but amusing pun there!)
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    Yes indeed 0.9rec = 0 + 9/10*(1+1/10+1/10^2....+1/10^n) as n tends to infinity

    You can recognise the sum of a geometric serie:

    1+1/10+ 1/10^2.....+1/10^n = 1/(1-1/10) = 10/9 (as n tends to infinity)

    Hence 0.9rec = 9/10*10/9 = 1 qed
 
 
 
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