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

Expand Messages
• 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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[Non-text portions of this message have been removed]
• 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]
• 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

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]
>
• 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:

>
> 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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• 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:

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:
>
> >
> > 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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• 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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