Channel Talk

# Re: [Channel-talkalgebra] Web Sites

From: Mrs. Dudones <cf_dudones@cfalls.org>
Date: Mon Apr 30 2007 - 17:41:09 EDT
X-Mailer: Microsoft Office Outlook 11

Thanks!

-----Original Message-----
From: channel-talkalgebra-bounces@learner.org
[mailto:channel-talkalgebra-bounces@learner.org] On Behalf Of
askey@math.wisc.edu
Sent: Monday, April 30, 2007 1:55 PM
To: channel-talkalgebra@learner.org
Cc: askey@math.wisc.edu
Subject: Re: [Channel-talkalgebra] Web Sites

I can recommend a book which goes well beyond what you will do,
but the first 35 or so pages out of 149 have a lot which any
middle school teacher who is starting the transition from
arithmetic to algebra will find useful.

The book is "Algebra" by I.M. Gelfand and A. Shen. It is
published by Birkhauser-Boston

Here is part of their explanation of the use of letters.

Problem 37 A small vessel and a big vessel contain (together)
5 liters. Two small and three big vessels contain together
13 liters. What are the volumes of the two vessels?

The first solution is "arithmetic". The small and the big
vessels together contin 5 liters. Therefore, two small vessels
and two large vessels contain 10 liters (10=2*5). As you know,
two small vessels and three big vessels contain 13 liters. So
we get 13 liters instead of 10 by adding one big vessel.
Therefore the volume of a big vessel is 3 liters. Now it is
easy to find the volume of a small vessel: together they
contain 5 liters, so a small vessel contains 5 - 3 = 2 liters.

This solution can be shortened if we use Vol.SV" instead of
"Volume of a Small Vessel" and "Vol.BV" instead of "Volume of
a Big Vessel." Thus, according to the statement of the problem,

Vol.SV + Vol.BV = 5,

therefore

2*Vol.SV + 2*Vol.BV = 10.

We also know that

2*Vol.SV + 3*Vol.BV = 13.

If we subtract the preceding equality from the last one we find
that Vol.BV = 3. Now the first equality implies that Vol.SV = 2.

Now the only thing to do is to replace our "Vol.SV" and "Vol.BV"
by standard unknowns x and y - and we get the standard "algebraic"
solution of our problem.

[You can fill in the remaining part of this paragraph.]

This is followed by a different use of a letter.

"Magic trick". Choose any number you wish. Add 3 to it. Multiply
the resut by 2. Subtract the chien number. Subtract 4. Subtract
the chosen number once more. You get 2, don't you?

Problem 38. Explain why this trick is successful.

Dick Askey

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Received on Tue May 1 08:56:04 2007

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