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I found a math breakthough .I think

  1. Nov 18, 2004 #1
    I found a math breakthough.....I think

    I just found another way to square a number.

    Ex1: 12^2=144
    but then this also equal 144
    12/2=6 'Number divided by 2
    12*2=24 'Number multiplied by 2
    24*6=144 <---'Both number multiplied
    24/4=4 <---'Both number divided

    Ex2: 15^2=225
    but then this also equal 225
    15/2=7.5 'Number divided by 2
    15*2=30 'Number multiplied by 2
    30*7.5=225 <---'Both number multiplied
    30/7.5=4 <---'Both number divided

    Everytime you follow this thing I made up and the divide them, you always get 4.

    Please don't insult me if this is stupid because I know I am a idiot. :smile:
  2. jcsd
  3. Nov 18, 2004 #2


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    Replace your number with x. Work through the algebra and you will see why things work out the way they do.
  4. Nov 18, 2004 #3
    Hehe, I remember back in the day when I was in middle school and every other day I'd think I found a break through in math.

    I remember being pretty excited at figuring out that for small two digit numbers, when you multiply by 11, all you have to do is simply:
    place the first digit first
    add the two digits and then place them in the second place digit
    place the second digit last

    And voila! you get the answer. For example, 43 x 11 =4 (3+4) 3 = 473

    I also remember proving that for any two triangles:
    if two sides from the first triangle = two sides from the second triangle and the included angles between those two sides add up to 270 degrees, their areas were equal. Not using trig!!!

    Also tried to figure out other ways to find PI but not everything that I tried worked :tongue2:

    Hehe, the good old days :tongue:
    Last edited: Nov 18, 2004
  5. Nov 18, 2004 #4


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    Raza, just in case you don't realize what hurkyl said:

    [tex]( \frac{x}{2})(2x) = x^2[/tex]
  6. Nov 18, 2004 #5
    Where do you get formulas written like that?
    And I still don't get you always get the answer 4
  7. Nov 18, 2004 #6


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    You need to use latex code to write formulas like that. If you click on any equation written in latex, it shows the code used to type it. There should be a link to the thread in the site FAQ.

    [tex] \frac{2x}{\frac{x}{2}} [/tex]

    multiply top and bottom by 2:

    [tex] \frac{4x}{x} = 4 [/tex]

    since the x's cancel
  8. Nov 19, 2004 #7


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    I love finding things like this as well. Such as the square of any number is the the two number around it multiplied plus 1. (x-1)(x+1) + 1 = x^2 which is because (x-y)(x+y) = x^2 - y^2

    I also found my own equation for pi involving dividing by tan(180/x) or something... but I can't seem to remember it!

    Nothing annoyed me more than when I realized that dx/dy(x^y) = yx^(y-1) was already found. :frown:
  9. Nov 19, 2004 #8


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    These 'rules' are really useful when doing arithmetic by head (what's it called in english?), you know, without a calculator.

    I used to suck at doing calculations in my head (actually I still do. My short-term memory just leaves at that moment), but applying all these tricks really helped me alot.

    For example: using (x-y)(x+y) = x^2 - y^2.
    15*17 (looks tough to me) = (16+1)(16-1)=16^2-1=255.
    You'll have to learn the squares from 1^2 to 20^2 by heart. By doing so you can make lots of calculations easier.
    This last can be used in conjunction with a trick for squaring a number that ends with 5: Take the number in front of the last 5 and multiply it with that number+1, then glue 25 to the result.
    i.e. 15*15=(1)(2) '+' 25=225
    65*65=(6)(7) '+' 25 = 4225
    Proof: take any number x that ends in 5. Call the number before the 5 d, so that x=10d+5

    Some of my math teachers can multiply two random two-digit numbers effortlessly. I wonder if they are just arithmetic wonders, or use a lot of these tricks...
    I never had such a great feel for numbers though.
    Okay, I`m going off topic..
    Last edited: Nov 19, 2004
  10. Nov 19, 2004 #9


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    I can multiply two digit numbers in my head faster than your average person using a piece of paper.
  11. Nov 19, 2004 #10


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    Do a web search on a fellow called Trachtenberg. BTW this works for all multiplications by 11.
    Last edited: Nov 19, 2004
  12. Nov 19, 2004 #11


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    Also, an integer is divisible by 11 if the difference between the sum of its odd-position digits and its even-position digits is divisible by 11. E.g. 192918 is divisible by 11 because 9+9+8-(1+2+1)=22
  13. Nov 21, 2004 #12
    I don't know why you multiply by 2

    Isn't it like this?

    [tex]( \frac{x}{2})(2x)[/tex]




  14. Nov 21, 2004 #13
    [tex](\frac{x}{2})(\frac{4x}{2}) = \frac{4x^2}{4}=x^2[/tex]
  15. Nov 21, 2004 #14


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    When multiplying fractions, you multiply numerators as well as denominators.
  16. Nov 21, 2004 #15
    If there's one thing I've learned it's that someone has always done something or thought about something well before you. I remember being so happy when I found a relationship between Pascal's Triangle and n-dimensional objects made up of one type of convex polygon. I even did a project on it and placed in the state science fair and nobody told me that it had been done before. I would have gone to ISEF except I had incredibly bad luck and had to compete for a spot in it with an RSI girl... but enough about my past :p. It wasn't until a few months later when I found some website stuck way back in the annals of the web that touched on the subject. Moral of the story: Unless you link two VERY esoteric parts of math, chances are it has been conceived and beaten into the ground.
  17. Nov 23, 2004 #16


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    If you learn to square two digit numbers in your head, you don't have to memorize any.

    Any 2 digit decimal number can be converted to Roman Numerals (don't worry, this is a way of thinking about numbers, so you don't actually have to write in Roman Numerals).

    36 is XXXVI - you have 3 tens, 1 five, and 1 one.
    39 is IXXXX - you have 4 tens and -1 ones.
    44 is XXXXIV - you have 4 tens, 1 five, and -1 ones.

    The idea is that it's simpler to multiply by tens and fives and then adjust to the precise value. If you modify the Roman Numeral System you learned in school:

    33 is XXXIIV - you have 3 tens, 1 five, and -2 ones.

    Now you can only have 5 different values for your ones: -2, -1, 0, 1, or 2.

    Using that principal, you can break any two digit number into an x + y + z format.

    x will be your tens, y will be your fives (it can only be 0 or 5), and z will be your ones (-2, -1, 0, 1, or 2).

    Squaring the sum of (x + y + z) is:
    [tex]x^2 + 2xy + y^2 + 2xz + 2yz + z^2[/tex]

    You can rearrange the second term via commutative property. The fourth and fifth term have 2z in common, so you can use the distributive property to get:

    [tex]x^2 + x(2y) + y^2 + (x+y)*2*z + z^2[/tex]

    Now, plug in your numbers.

    The tens are squared. That's easy. You squaring a one digit number and doubling the number of zeros.

    Multiply x by (2 * y). 2*y will either equal 10 (if y is 5) or 0 (if y is 0). Easy step. x is multiplied by 10 or 0. Add this to your first step.

    Square y. You get either 25 (y is 5) or 0 (y is 0). Add this to the sum from your second step.

    The fourth term is done exactly as it is written: 1) add x + y 2)multiply by 2 3) multiply by z (When you're doing this in your head, it's easier to multiply by 2 twice than it is to multiply by 4 once). Add this to the sum from your third step (z might be negative, so this is the only tough step in the whole procedure).

    Square z. Add to the sum from your fourth step.

    Example: 36 = 30 + 5 + 1, so:

    30^2 is 900
    10*30 is 300 giving you 1200, so far.
    5^2 is 25, giving 1225, so far.
    35 * 2 * 1 is 70, giving you 1295, so far.
    1^2 is 1, giving you your final answer of 1296.

    33 = 30 + 5 - 2, so:

    30^2 is 900
    30*10 is 300, giving you 1200.
    5^2 gets you to 1225.
    35 * 2 * -2 is -140 (or -200 + 60) giving you 1085.
    (-2)^2 is 4, giving you 1089.

    35 = 30 + 5 + 0
    30^2 is 900
    30*10 gets you to 1200.
    5^2 gets you to 1225.
    Both the last two steps involve multiplication by 0, so they can be deleted.
  18. Nov 23, 2004 #17


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    That's a cool post BobG. Thanks, I hope it will be of help.

    I thought 3 was written III instead of IIV. Nevermind though, it doesn't relate to the calculation anyway.

    I guess you'll just have to memorize the steps then (and lots of practice ofcourse).
    The problem I have is that when there are lots of steps involved I forget what I have done, but since these steps are all easy, the chances of not forgetting increases.
  19. Nov 23, 2004 #18


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    It is. But if you use it as normally written, squaring numbers that end in 3 or 8 are a royal pain. -2 is a lot better number.
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