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Re: [Mental Calculation] Deriving the Quarter Squares-Rule.

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  • Andy Robertshaw
    Not too difficult really. ((a+b)/2)^2 = (a+b)/2 * (a+b)/2 = ((a+b)^2)/4 = (a^2 +2ab + b^2)/4 and ((a-b)/2)^2= (a-b)/2 * (a-b)/2 = ((a-b)^2)/4 = (a^2 -2ab +
    Message 1 of 8 , Jan 31, 2007
      Not too difficult really.

      ((a+b)/2)^2 = (a+b)/2 * (a+b)/2 = ((a+b)^2)/4 = (a^2 +2ab + b^2)/4
      and
      ((a-b)/2)^2= (a-b)/2 * (a-b)/2 = ((a-b)^2)/4 = (a^2 -2ab + b^2)/4

      Therefore ((a+b)/2)^2 - ((a-b)/2)^2 = (a^2 +2ab + b^2)/4 - (a^2 -2ab + b^2)/4 =4ab/4 = ab

      Hope this helps,
      Andy



      ----- Original Message ----
      From: peterschreyfogl <peterschreyfogl@...>
      To: MentalCalculation@yahoogroups.com
      Sent: Tuesday, 30 January, 2007 8:02:59 PM
      Subject: [Mental Calculation] Deriving the Quarter Squares-Rule.

      Hello!

      At http://de.wikipedia .org/wiki/ Kopfrechnen# Rechentricks it is stated
      that one can derive the Quarter Squares-Rule by transforming the
      Binomials 1. (a + b)^2 = a^2 + 2ab + b^2 and 2. (a - b)^2 = a^2 - 2ab
      + b^2.

      Well, I get to the point where: ab = ((a + b)^2 - a^2 - b^2)/2, but
      how do I get to ab = ((a + b)/2)^2 - ((a - b)/2)^2?

      y.s.,
      Michael (peter.schreyfogl)






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    • Ian Docherty
      Hmm. this sounds like a homework question... anyway, a hint to get you started you have ab = ((a + b)^2 - a^2 - b^2)/2 hence ab = ((a + b)^2)/2 - (a^2 + b^2)/2
      Message 2 of 8 , Jan 31, 2007
        Hmm. this sounds like a homework question...

        anyway, a hint to get you started

        you have ab = ((a + b)^2 - a^2 - b^2)/2
        hence ab = ((a + b)^2)/2 - (a^2 + b^2)/2 ...(1)

        From (a - b)^2 = a^2 - 2ab + b^2
        you can re-arrange to get a value for (a^2 + b^2) and substitute in the
        equation ...(1) above

        The rest is simple algebra....

        Regards
        Ian


        peterschreyfogl wrote:
        >
        > Hello!
        >
        > At http://de.wikipedia.org/wiki/Kopfrechnen#Rechentricks
        > <http://de.wikipedia.org/wiki/Kopfrechnen#Rechentricks> it is stated
        > that one can derive the Quarter Squares-Rule by transforming the
        > Binomials 1. (a + b)^2 = a^2 + 2ab + b^2 and 2. (a - b)^2 = a^2 - 2ab
        > + b^2.
        >
        > Well, I get to the point where: ab = ((a + b)^2 - a^2 - b^2)/2, but
        > how do I get to ab = ((a + b)/2)^2 - ((a - b)/2)^2?
        >
        > y.s.,
        > Michael (peter.schreyfogl)
        >
        > __._,_.



        [Non-text portions of this message have been removed]
      • peterschreyfogl
        Thank s for your reply, Scott. I see. It is the proof that the Quarter Squares-Rule produces the result ab. Do you see also how one can derive the identity by
        Message 3 of 8 , Feb 1, 2007
          Thank's for your reply, Scott.

          I see. It is the proof that the Quarter Squares-Rule produces the
          result ab.

          Do you see also how one can derive the identity by transforming the
          Binomial formulas? Maybe it is not correct like it is said on
          Wikipedia. Nonetheless, through some logic there are ways how one can
          get to the Quarter Squares-Rule. -- The question is HOW.

          t.y., Michael


          --- In MentalCalculation@yahoogroups.com, "P Scott Horne"
          <patrick.horne@...> wrote:
          >
          > ((a + b)/2)^2 = (a^2 + 2ab + b^2)/4
          >
          > ((a - b)/2)^2 = (a^2 - 2ab + b^2)/4
          >
          > Subtract the latter from the former, and you get
          >
          > (2ab + 2ab)/4,
          >
          > which is
          >
          > ab.
          >
          > QED.
          >
          > P Scott Horne
          >
          >
          >
          >
          > _____
          >
          > From: MentalCalculation@yahoogroups.com
          > [mailto:MentalCalculation@yahoogroups.com] On Behalf Of
          peterschreyfogl
          > Sent: 30 janvier 2007 15:03
          > To: MentalCalculation@yahoogroups.com
          > Subject: [Mental Calculation] Deriving the Quarter Squares-Rule.
          >
          >
          >
          >
          > Hello!
          >
          > At http://de.wikipedia
          > <http://de.wikipedia.org/wiki/Kopfrechnen#Rechentricks>
          > .org/wiki/Kopfrechnen#Rechentricks it is stated
          > that one can derive the Quarter Squares-Rule by transforming the
          > Binomials 1. (a + b)^2 = a^2 + 2ab + b^2 and 2. (a - b)^2 = a^2 -
          2ab
          > + b^2.
          >
          > Well, I get to the point where: ab = ((a + b)^2 - a^2 - b^2)/2, but
          > how do I get to ab = ((a + b)/2)^2 - ((a - b)/2)^2?
          >
          > y.s.,
          > Michael (peter.schreyfogl)
          >
          >
          >
          >
          >
          >
          >
          > [Non-text portions of this message have been removed]
          >
        • rob f
          Unlike the previous posts, my proof is algebra-free. Just stare at this picture long enough and the rule will appear to you stereogram-style . Arrange 4
          Message 4 of 8 , Feb 7, 2007
            Unlike the previous posts, my proof is algebra-free.

            Just stare at this picture long enough and the rule
            will appear to you "stereogram-style".

            Arrange 4 identical rectangles, each of dimension a x
            b, into a square, like this:

            * * * * * * * * * * * * * *
            * * *
            * * *
            * * *
            * * * * * * * * * * *
            * * * *
            * * * *
            * * * *
            * * * *
            * * * * * * * * * * *
            * * *
            * * *
            * * *
            * * * * * * * * * * * * * *

            (The above picture might get messed up when I send
            this, but I think it's fairly obvious what is meant)

            If you can't see the "quarter squares rule" in the
            picture then these observations might help:

            1. Area of large square = (a+b)^2

            2. Area of square "hole" in centre = (a-b)^2

            3. Area of large square = combined area of the four
            rectangles plus area of the square hole at centre

            Hence, (a+b)^2 = 4ab + (a-b)^2
            Divide through by 4 gives the rule.

            RobF


            --- peterschreyfogl <peterschreyfogl@...> wrote:

            > Thank's for your reply, Scott.
            >
            > I see. It is the proof that the Quarter Squares-Rule
            > produces the
            > result ab.
            >
            > Do you see also how one can derive the identity by
            > transforming the
            > Binomial formulas? Maybe it is not correct like it
            > is said on
            > Wikipedia. Nonetheless, through some logic there are
            > ways how one can
            > get to the Quarter Squares-Rule. -- The question is
            > HOW.
            >
            > t.y., Michael
            >
            >
            > --- In MentalCalculation@yahoogroups.com, "P Scott
            > Horne"
            > <patrick.horne@...> wrote:
            > >
            > > ((a + b)/2)^2 = (a^2 + 2ab + b^2)/4
            > >
            > > ((a - b)/2)^2 = (a^2 - 2ab + b^2)/4
            > >
            > > Subtract the latter from the former, and you get
            > >
            > > (2ab + 2ab)/4,
            > >
            > > which is
            > >
            > > ab.
            > >
            > > QED.
            > >
            > > P Scott Horne
            > >
            > >
            > >
            > >
            > > _____
            > >
            > > From: MentalCalculation@yahoogroups.com
            > > [mailto:MentalCalculation@yahoogroups.com] On
            > Behalf Of
            > peterschreyfogl
            > > Sent: 30 janvier 2007 15:03
            > > To: MentalCalculation@yahoogroups.com
            > > Subject: [Mental Calculation] Deriving the Quarter
            > Squares-Rule.
            > >
            > >
            > >
            > >
            > > Hello!
            > >
            > > At http://de.wikipedia
            > >
            >
            <http://de.wikipedia.org/wiki/Kopfrechnen#Rechentricks>
            > > .org/wiki/Kopfrechnen#Rechentricks it is stated
            > > that one can derive the Quarter Squares-Rule by
            > transforming the
            > > Binomials 1. (a + b)^2 = a^2 + 2ab + b^2 and 2. (a
            > - b)^2 = a^2 -
            > 2ab
            > > + b^2.
            > >
            > > Well, I get to the point where: ab = ((a + b)^2 -
            > a^2 - b^2)/2, but
            > > how do I get to ab = ((a + b)/2)^2 - ((a -
            > b)/2)^2?
            > >
            > > y.s.,
            > > Michael (peter.schreyfogl)
            > >
            > >
            > >
            > >
            > >
            > >
            > >
            > > [Non-text portions of this message have been
            > removed]
            > >
            >
            >
            >






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          • rob f
            A I expected, the picture got messed up. I would upload a picture-file to the group, but that feature seems to have been disabled. I place here instead:
            Message 5 of 8 , Feb 7, 2007
              A I expected, the picture got messed up. I would
              upload a picture-file to the group, but that feature
              seems to have been disabled. I place here instead:

              http://www.mentalcalculation.com/misc/quad_rule_picture.pdf

              The 4 rectangles should be the same size, but I'm no
              artist...

              Rob F

              --- rob f <rob221b@...> wrote:

              > Unlike the previous posts, my proof is algebra-free.
              >
              > Just stare at this picture long enough and the rule
              > will appear to you "stereogram-style".
              >
              > Arrange 4 identical rectangles, each of dimension a
              > x
              > b, into a square, like this:
              >
              > * * * * * * * * * * * * * *
              > * * *
              > * * *
              > * * *
              > * * * * * * * * * * *
              > * * * *
              > * * * *
              > * * * *
              > * * * *
              > * * * * * * * * * * *
              > * * *
              > * * *
              > * * *
              > * * * * * * * * * * * * * *
              >
              > (The above picture might get messed up when I send
              > this, but I think it's fairly obvious what is meant)
              >
              > If you can't see the "quarter squares rule" in the
              > picture then these observations might help:
              >
              > 1. Area of large square = (a+b)^2
              >
              > 2. Area of square "hole" in centre = (a-b)^2
              >
              > 3. Area of large square = combined area of the four
              > rectangles plus area of the square hole at centre
              >
              > Hence, (a+b)^2 = 4ab + (a-b)^2
              > Divide through by 4 gives the rule.
              >
              > RobF
              >
              >
              > --- peterschreyfogl <peterschreyfogl@...>
              > wrote:
              >
              > > Thank's for your reply, Scott.
              > >
              > > I see. It is the proof that the Quarter
              > Squares-Rule
              > > produces the
              > > result ab.
              > >
              > > Do you see also how one can derive the identity by
              > > transforming the
              > > Binomial formulas? Maybe it is not correct like it
              > > is said on
              > > Wikipedia. Nonetheless, through some logic there
              > are
              > > ways how one can
              > > get to the Quarter Squares-Rule. -- The question
              > is
              > > HOW.
              > >
              > > t.y., Michael
              > >
              > >
              > > --- In MentalCalculation@yahoogroups.com, "P Scott
              > > Horne"
              > > <patrick.horne@...> wrote:
              > > >
              > > > ((a + b)/2)^2 = (a^2 + 2ab + b^2)/4
              > > >
              > > > ((a - b)/2)^2 = (a^2 - 2ab + b^2)/4
              > > >
              > > > Subtract the latter from the former, and you get
              > > >
              > > > (2ab + 2ab)/4,
              > > >
              > > > which is
              > > >
              > > > ab.
              > > >
              > > > QED.
              > > >
              > > > P Scott Horne
              > > >
              > > >
              > > >
              > > >
              > > > _____
              > > >
              > > > From: MentalCalculation@yahoogroups.com
              > > > [mailto:MentalCalculation@yahoogroups.com] On
              > > Behalf Of
              > > peterschreyfogl
              > > > Sent: 30 janvier 2007 15:03
              > > > To: MentalCalculation@yahoogroups.com
              > > > Subject: [Mental Calculation] Deriving the
              > Quarter
              > > Squares-Rule.
              > > >
              > > >
              > > >
              > > >
              > > > Hello!
              > > >
              > > > At http://de.wikipedia
              > > >
              > >
              >
              <http://de.wikipedia.org/wiki/Kopfrechnen#Rechentricks>
              > > > .org/wiki/Kopfrechnen#Rechentricks it is stated
              > > > that one can derive the Quarter Squares-Rule by
              > > transforming the
              > > > Binomials 1. (a + b)^2 = a^2 + 2ab + b^2 and 2.
              > (a
              > > - b)^2 = a^2 -
              > > 2ab
              > > > + b^2.
              > > >
              > > > Well, I get to the point where: ab = ((a + b)^2
              > -
              > > a^2 - b^2)/2, but
              > > > how do I get to ab = ((a + b)/2)^2 - ((a -
              > > b)/2)^2?
              > > >
              > > > y.s.,
              > > > Michael (peter.schreyfogl)
              > > >
              > > >
              > > >
              > > >
              > > >
              > > >
              > > >
              > > > [Non-text portions of this message have been
              > > removed]
              > > >
              > >
              > >
              > >
              >
              >
              >
              >
              >
              >
              >
              ___________________________________________________________
              >
              > New Yahoo! Mail is the ultimate force in competitive
              > emailing. Find out more at the Yahoo! Mail
              > Championships. Plus: play games and win prizes.
              >
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              >
              >




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            • peterschreyfogl
              Hello Rob! Yes, this is it. You could also say 4ab = (a+b)^2 - (a-b)^2 (0) And (a+b)^2 = a^2 + 2ab + b^2 (1) is the 1. Binomische Formel (a-b)^2 = a^2 - 2ab
              Message 6 of 8 , Feb 7, 2007
                Hello Rob!

                Yes, this is it. You could also say

                4ab = (a+b)^2 - (a-b)^2 (0)

                And

                (a+b)^2 = a^2 + 2ab + b^2 (1) is the 1. Binomische Formel
                (a-b)^2 = a^2 - 2ab + b^2 (2) is the 2. Binomische Formel

                So, 4ab = left or right part of (1) - left or right part of (2),
                which will lead to equation (0).

                If we divide equation (0) by 4, we get

                ab = ((a+b)^2 - (a-b)^2) / 4

                is equivalent to

                ab = (a+b)^2/4 - (a-b)^2/4

                and this is equivalent to

                ab = ((a+b)/2)^2 - ((a-b)/2)^2

                which is the Quarter Squares Rule. []


                ----

                Do you know how "1. Binoische Formel" is called in English?

                ----~~ Peter Schreyfogl

                --- In MentalCalculation@yahoogroups.com, rob f <rob221b@...> wrote:
                >
                >
                > --- rob f <rob221b@...> wrote:
                >
                > ...
                > >
                > > If you can't see the "quarter squares rule" in the
                > > picture then these observations might help:
                > >
                > > 1. Area of large square = (a+b)^2
                > >
                > > 2. Area of square "hole" in centre = (a-b)^2
                > >
                > > 3. Area of large square = combined area of the four
                > > rectangles plus area of the square hole at centre
                > >
                > > Hence, (a+b)^2 = 4ab + (a-b)^2
                > > Divide through by 4 gives the rule.
                > >
                > > RobF
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