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Solutions for Session 9, Part H
See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6| H7| H8| H9| H10| H11
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Problem H1 | |
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a:
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Step 1: | YZYZZYYZ |
Step 2: | YZYYYZ (erase ZZ) |
Step 3: | YZZ (erase YYY) |
Step 4: | Y (erase ZZ) |
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b:
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Step 1: | YYYYZZYZY |
Step 2: | YZZYZY (erase YYY) |
Step 3: | YYZY (erase ZZ) |
Step 4: | YYYZ (commute last ZY) |
Step 5: | Z (erase YYY) |
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c:
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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) |
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<< back to Problem H1
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Problem H2 | |
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The elements of the YZ group are E, Y, YY, Z, YZ, and YYZ.
<< back to Problem H2
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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
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Problem H5 | |
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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
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Problem H6 | |
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Since Y3 is identical to E, Y4 will be identical to Y, Y5 = YY, Y6 = E, etc.
<< back to Problem H6
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Problem H7 | |
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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
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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
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Problem H9 | |
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Here is the completed table:
Note that every element appears exactly once in each row, and once in each column.
<< back to Problem H9
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Problem H10 | |
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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
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Problem H11 | |
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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
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