Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Monthly Update sign up
Mailing List signup
Search
MENU
Learning Math Home
Patterns, Functions, and Algebra
 
Session 9 Part A Part B Part C Part D Part E Homework
 
Glossary
Algebra Site Map
Session 9 Materials:
Notes
Solutions
 

A B C D
Homework

Video

Solutions for Session 9, Part H

See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6| H7| H8| H9| H10| H11


Problem H1

a:
 

Step 1:

YZYZZYYZ

Step 2:

YZYYYZ (erase ZZ)

Step 3:

YZZ (erase YYY)

Step 4:

Y (erase ZZ)

b:
 

Step 1:

YYYYZZYZY

Step 2:

YZZYZY (erase YYY)

Step 3:

YYZY (erase ZZ)

Step 4:

YYYZ (commute last ZY)

Step 5:

Z (erase YYY)

c:
 

Step 1:

YZYZYZYZYZYZYZYZZZYZYZYYZY (better think more systematically)

Step 2:

YYYYYYYYYYYYYZZZZZZZZZZZZZ (commute all Ys first, Zs last)

Step 3:

YZ (erase 12 Ys by threes, 12 Zs by twos)

<< back to Problem H1


 

Problem H2

The elements of the YZ group are E, Y, YY, Z, YZ, and YYZ.

<< back to Problem H2


 

Problem H3

a. 

E * YZ = YZ

b. 

YZ * YY = Z

c. 

Z * YZ = Y

Note that in all of these, any three occurrences of Y can be removed, as can any two occurrences of Z. Since the commutative law exists for this group, order is not important.

<< back to Problem H3


 

Problem H4

a. 

YZ * YYZ = E

b. 

Z * Y = YZ

c. 

YY * YZ = Z

<< back to Problem H4


 

Problem H5

Y3 means the same as Y * Y * Y, which is YYY, which is the same as E. The same is true of Z2, which is identical to Z * Z.

<< back to Problem H5


 

Problem H6

Since Y3 is identical to E, Y4 will be identical to Y, Y5 = YY, Y6 = E, etc.

<< back to Problem H6


 

Problem H7

Powers of each element:

 

E: E only

 

Y: Y1 = Y, Y2 = YY, Y3 = E

 

YY: YY1 = YY, YY2 = Y, Y3 = E

 

Z: Z1 = Z, Z2 = E

 

YZ: YZ1 = YZ, YZ2 = YY, YZ3 = Z, YZ4 = Y, YZ5 = YYZ, YZ6 = E

 

YYZ: YYZ1 = YYZ, YYZ2 = Y, YYZ3 = Z, YYZ4 = YY, YYZ5 = YZ, YYZ6 = E

<< back to Problem H7


 

Problem H8

a. 

Y1,000 = Y1, since 1,000 = 1 (mod 3), and the powers of Y repeat every three powers.

b. 

(YZ)1,001 = (YZ)5, since 1,001 = 5 (mod 6), and the powers of YZ repeat every six powers. According to the list of powers of YZ, (YZ)5 = YYZ. Another way to do this is to imagine a line of 1,001 Ys and 1,001 Zs, and decide what would be left after all the cancellation.

<< back to Problem H8


 

Problem H9

Here is the completed table:

yz table

Note that every element appears exactly once in each row, and once in each column.

<< back to Problem H9


 

Problem H10

The element E works this way, since E * A = A for any element A in the table, just like 1 * N = N for any number N.

<< back to Problem H10


 

Problem H11

The reciprocals can be found by finding E within the row and column of each element. Here are the reciprocals, in pairs:

 

E and E

 

Y and YY

 

Z and Z

 

YZ and YYZ

<< back to Problem H11

 

Learning Math Home | Algebra Home | Glossary | Map | ©

Session 9: Index | Notes | Solutions | Video

Home | Catalog | About Us | Search | Contact Us | Site Map

  • Follow The Annenberg Learner on Facebook

© Annenberg Foundation 2013. All rights reserved. Privacy Policy