Download Midterm: model solutions and mark scheme File

MTH4104
Introduction to Algebra
In-term test: model solutions and mark scheme
Semester B, 2017
Question 1 (20 marks) Prove by mathematical induction that, for any integer n > 1,
n
∏ (22
j−1
+ 1) = 22 − 1.
n
Solution Let P(n) be the statement “∏nj=1 (22
j−1
+ 1) = 22 − 1”.
j=1
0
n
1
Base case: P(1) says 22 + 1 = 22 − 1 which is true, as both sides equal 3.
Inductive step: Suppose that P(k) is true for some integer k ≥ 1, i.e.
0
1
k−1
(22 + 1) · (22 + 1) · · · (22
k
+ 1) = 22 − 1.
We have to deduce P(k + 1) from this, which says
0
1
k−1
(22 + 1) · (22 + 1) · · · (22
k
k+1
+ 1) · (22 + 1) = 22
− 1.
Now the inductive hypothesis allows us to substitute for most of the terms on the left side:
0
1
k−1
(22 + 1) · (22 + 1) · · · (22
k
k
k
+ 1) · (22 + 1) = (22 − 1) · (22 + 1),
and multiplying this out yields
k
k
k+1
(22 )2 − 1 = 22·2 − 1 = 22
−1
by laws of exponents. This equals the right hand side of P(k + 1), so we have proved that P(k) implies
P(k + 1) for any integer k ≥ 1.
Conclusion: Therefore by induction, P(n) is true for all integers n ≥ 1.
Common errors
1A Replacing multiplication by addition. The sign ∏ means that successive terms are to be multiplied, not
added like ∑.
1B Writing the proof of equality by manipulating both sides of P(k +1) until you get two equal things. This
is logically incorrect: proving a true statement having assumed P(k + 1) is not the same as proving
P(k + 1). It’s usually possible to convert this to an actual proof easily: rather than writing a chain of
equalities, compute the LHS and RHS separately and find that they are equal.
k
1C Mishandling the nested exponents. There is no law of exponents that lets you collapse 22 to anything
simpler; maybe you were confusing it with (22 )k = 22·k ?
1
n
1D Ignoring, or losing, the “ ∏ ” sign and using just one term on the left hand side. ∏ is a notation for
j=1
products, just like ∑ is a notation for sums:
n
∏ f ( j)
means
f (1) f (2) · · · f (n).
j=1
1E When substituting in k or k + 1 for n to get the statement of P(k) or P(k + 1), you also substituted for j.
But j is different to n; the variable j is just a “dummy” variable in the sum and should be left alone.
Observe the difference between
k
∏ f ( j) = f (1) f (2) · · · f (k − 1) f (k)
j=1
and
k
∏ f ( j) = f (k) f (k) · · · f (k) f (k).
j=1
Rubric Structure of an inductive proof, i.e. stating P(n) and having a base case and an inductive step: 4
marks altogether.
• Separating n = 1 and another case: 2 marks
• Indicating that P(k) is to be used to prove P(k + 1) in the inductive step: 2 marks
Proof of base case: 3 marks.
Proof of inductive step: 13 marks altogether.
• Correct statement of P(k + 1): 2 marks.
• Substituting the equation from P(k) into the product in P(k + 1): 7 marks, of which 2 awarded for
substituting in an unuseful way
• Proving P(k + 1) from there: 4 marks.
“Innocuous” errors in manipulation −2 marks, in notation −1 mark.
(But if you just say two things are equal because they need to be to finish the proof, and they aren’t, that’s
not an innocuous error.)
Question 2 (6+7+7 marks) (a) Give definitions of the set of hypercomplex numbers and of its addition
and multiplication operations.
(b) Prove or disprove: the pseudocomplex numbers satisfy the multiplicative inverse law.
(c) Prove or disprove: the quaternions satisfy the distributive law.
Solution (a) The set of hyper-complex numbers is defined to be
B := {α + β j : α, β ∈ C}.
We define addition and multiplication of hyper-complex numbers by the formulae
(α + β j) + (γ + δ j) := (α + γ) + (β + δ ) j,
(α + β j)(γ + δ j) := (αγ − β δ ) + (αδ + β γ) j.
(b) This assertion is false. To disprove it, we will show that the pseudocomplex number 1 +t is a counterexample to the law, having no multiplicative inverse. First, we compute
(1 − t)(1 + t) = 1 − t 2 = 0.
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Suppose for a contradiction that u := 1 + t has a multiplicative inverse v. Multiply the preceding equation
by v. Using the associative and identity laws and the fact that uv = 1 we obtain
0 = 0v = ((1 − t)(1 + t)) v = (1 − t) ((1 + t)v) = (1 − t)(uv) = (1 − t)1 = 1 − t.
When written in standard form, 0 = 0 + 0t, so 0 + 0t = 1 + (−1)t. Equating the real parts gives the absurd
equation 0 = 1, a contradiction. We have therefore
(c) This assertion is true. To prove it, we let a + b j, c + d j, e + f j be three arbitrary
quaternions, where
a, b, c, d, e, f are complex numbers. We must prove that (a + b j) (c + d j) · (e + f j) and (a + b j)(c + d j) +
(a + b j)(e + f j) are equal. Working these out using the definitions, the first is
(a + b j) (c + d j) · (e + f j) = (a + b j) (c + e) + (d + f ) j
= a(c + e) − b(d + f ) + a(d + f ) + b(c + e) j.
The second is
(a + b j)(c + d j) + (a + b j)(e + f j) = (ac − bd) + (ad + bc) j + (ae − b f ) + (a f + be) j
= (ac − bd + ae − b f ) + (ad + bc + a f + be) j.
These are equal componentwise, once we use the rule z + w = z + w and expand out the product. This
proves the distributive law.
Common errors
(name of a law) Using the wrong law. If you did this I’ve written down the name of the law you were
using.
(name of a number system) Defining the wrong number system, similarly.
2A In part (b), showing that one number b is not the inverse of a given number a does not imply that a has
no inverse.
2B A definition should ideally contain just the term or expression thing being defined, and what it means.
You have included working in your sum and/or product formulas. This working is what motivates our
definition, but it doesn’t actually belong there.
2C In part (b), you have showed that one number a has an inverse. But you have to treat all possible
nonzero numbers.
2D Your proof of the distributive law involved expanding parentheses in a product with j involved. But it
is the distributive law that lets you do that! So your argument is circular.
2E Watch out for division by zero in your formula for the multiplicative inverse.
2F Insufficiently general element. When choosing variables for the components of some number (real
and imaginary components, say), you have accidentally left some variables out, and therefore your
argument didn’t handle every number. This can also happen if you reuse letters, e.g. you use a + bi
for two different complex numbers.
2G Not giving a concrete counterexample when you said a law was false. To be 100% sure, we want an
actual number, for example 1 + t, not something with variables still in the coefficients.
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Rubric For part (a): 2 marks for each of the set, the sum, and the product.
For parts (b) and (c):
Making the correct choice whether to prove or disprove: 3 marks.
Argument: the remaining 4 marks.
Question 3 (6+9+5 marks) (a) Let a and b be positive integers. Define the greatest common divisor and
the least common multiple of a and b.
(b) Use Euclid’s algorithm to compute the greatest common divisor of 288 and 102.
(c) What is the least common multiple of 288 and 102? (It is OK if you give an answer with unevaluated
arithmetic operations left in it.)
Solution (a) We call a natural number d the greatest common divisor of a and b if it is a common divisor
of a and b, i.e. it divides them both; and any other common divisor of a and b is smaller than d.
The natural number m is the least common multiple of a and b if it is a common multiple of a and m,
i.e. both a and b divide it; and m is smaller than any other common multiple of a and b.
(b) The Euclidean algorithm consists of taking repeated remainders:
288 = 2 · 102 + 84
102 = 1 · 84 + 18
84 = 4 · 18 + 12
18 = 1 · 12 + 6
12 = 2 · 6 + 0.
So the gcd of 288 and 102 is seen to be 6.
(c) By one of the starred coursework problems from week 4, we know that gcd(a, b) · lcm(a, b) = a · b for
any natural numbers a and b. Therefore, lcm(288, 102) = 288 · 102/6.
Alternative solution, by prime factorisation: 288 = 25 32 and 102 = 2 · 3 · 17, so the lcm is 25 32 17.
These work out to 4896.
Common errors
3A In the definitions of gcd and lcm, we must specify that we are considering positive numbers. There is
no least common multiple of two integers if negative multiples are allowed; you could take a negative
common multiple of any magnitude.
3B You have given a procedure for computing the gcd (or part thereof), as opposed to its definition.
3C A definition should be concise and understandable by itself; you’ve given some extra chatter. An
example (say) following the definition is a good thing to include in a book, but it’s not part of the
definition. (And it takes up time on the test!)
3D The definition of the gcd (say) should indicate clearly what number the gcd is! For instance, don’t start
“The greatest common divisor is when d | a and d | b and. . . ”; instead, say “The greatest common
divisor is the number d such that . . . ”.
3E You used subtraction instead of mod to do Euclid’s algorithm. This works, though it’s not the way I
taught it, and it’s less efficient.
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Rubric Definition: 3 marks for each. Deductions for imprecision and uncareful writing.
Euclidean algorithm: 2 marks deducted per arithmetic error, 3 per algebraic error in workings, 5 for a
significant algorithmic error. Computing the gcd a different way than the question specified is worth 4
marks.
LCM: full marks for a correct number. Up to 2 marks for method alone. Deductions: -2 for errors in
factoring, -1 for more mechanical errors.
Question 4 (7+7+6 marks) (a) Let R ⊆ X 2 be a relation on a set X. Define fully what it means for R to be
an equivalence relation, including the definitions of any properties of relations you invoke.
(b) Prove that the relation
R = {(a, b) ∈ R2 : b − a is a natural number}
on the set of real numbers is transitive.
(c) Write down the partition of the set Y = {1, 2, 3, 4, 5, 6} into equivalence classes of the relation
S = {(x, y) ∈ Y 2 : x + y is even}.
You may assume S is an equivalence relation.
Solution (a) We say that R is
reflexive if (x, x) ∈ R for all x ∈ X;
symmetric if (x, y) ∈ R implies that (y, x) ∈ R;
transitive if (x, y) ∈ R and (y, z) ∈ R together imply that (x, z) ∈ R.
If R has all three of these properties, it is an equivalence relation.
(b) We must prove that for any three real numbers a, b, and c, if aRb and bRc then aRc. Let us assume aRb
and bRc. By definition of R, this implies that b − a = k and c − b = ` for some natural number k and `. By
adding these two equations together, one sees that c − a = k + `, which is a natural number. This is exactly
the content of aRc, so we have proved the transitive property.
(c) The partition is
{{1, 3, 5}, {2, 4, 6}}.
To find this, we pick an element of Y , say 1, and consider its equivalence class,
[1]S = {y ∈ Y : 1Sy} = {y ∈ Y : 1 + y is even.}
Clearly this is the set {1, 3, 5}. Now we do not have 2 in our partition yet, so we repeat the procedure:
[2]S = {y ∈ Y : 2Sy} = {y ∈ Y : 2 + y is even.}
This works out to {2, 4, 6}. And since the two equivalence classes we’ve found so far cover every element
of Y , we are done.
4A In part (b), providing a single example is not a proof. (In theory you could prove transitivity by checking
all possible triples by hand, but that would be an infinite amount of work.)
4B A partition of X is a set of subsets of X; its elements are equivalence classes, which are subsets of X. A
relation is a set of ordered pairs.
By using the wrong kind of brackets, you have mixed and matched these. Equivalence classes should
not have pairs or sets inside them, just numbers. Relations should not have sets {a, b} in them. Etc.
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4C If you have listed all the equivalence classes of S, you haven’t quite written down the partition according
to our definition of partition. To get the partition you have to put all the equivalence classes into a set.
4D In the proof in part (b), it matters that you get the definition of R the right way around, and write
b − a ∈ N and not a − b ∈ N (etc.) This is because the sum of two natural numbers is natural, but the
difference is not necessarily natural; it might be negative. That would stop the proof of transitivity
from working.
4E Confusion about the notation R2 or Y 2 . This means the Cartesian product: for example, (x, y) ∈ R2
means x ∈ R and y ∈ R. In particular, square numbers are not involved.
4F Extraneous material in the definition. See 3C above.
Rubric (a) 2 marks for each of the reflexive, symmetric, and transitive properties; 1 for the aggregate.
(b) 3 marks for reducing to “b − a ∈ N and c − b ∈ N implies c − a ∈ N”; 4 marks for proving that fact.
(c) 2 marks for listing S; 2 marks for the equivalence classes of S; 2 marks for assembling these into a
partition. (You need not show all the above intermediate steps; the marks will be awarded anyway.)
Question 5 (6+14 marks) (a) Let m be a positive integer. Give the definitions of the addition and multiplication operations on the set Zm .
(b) Find all solutions x ∈ Z11 to the equation [6]11 x + [1]11 =
[10]11
, with justification.
[6]11
Solution (a) The definitions are
[a]m + [b]m := [a + b]m ,
[a]m · [b]m := [ab]m .
(b) This problem was copied from the coursework! I will copy my solution from there as well. It is a
slightly roundabout one: the task is actually easier than the below if you clear the denominator first. . .
[10]11
The first thing to deal with is the expression
: we have to work out what that means. Division by
[6]11
[6]11 means multiplying by its multiplicative inverse; it has one, since it was proved in the Lectures that Z11
forms a field. By inspection, we spot the fact that
[2]11 · [6]11 = [12]11 = [1]11
so that [2]11 is the multiplicative inverse of [6]11 , and what we have on the right hand side of the equation is
[10]11 · ([6]11 )−1 = [10]11 · [2]11 = [20]11 = [9]11 .
Now, we can solve the linear equation
[6]11 x + [1]11 = [9]11
just as we always do:
[9]11 − [1]11 [8]11
=
= [8]11 · ([6]11 )−1 .
[6]11
[6]11
One more inverse calculation to do. Again we spot
x=
[6]11 · [2]11 = [12]11 = [1]11
so the inverse of [6]11 is [2]11 , and we can finish:
x = [8]11 · [2]11 = [16]11 = [5]11 .
Let’s check that this really works: 6 · 5 + 1 = 31 which is congruent to 9 modulo 11.
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Common errors
5A The answer x, and the coefficients in your workings, need to be elements of Z11 , not integers or rational
numbers. It’s true that the rules of algebra work the same way over one field as another, and that
computation in modular arithmetic is very similar to computation in the integers. That doesn’t mean
you can mix and match!
5B Simplify your answers down to a single element of Z11 ; don’t leave them as a ratio of two elements. I
have taught you procedures to do this, e.g. the extended Euclidean algorithm to invert the denominator.
5C The equivalence relation ≡11 is only defined for integers. We cannot take the equivalence class of a
fraction. Instead of, say, x = [1/9]11 , you should write x = [1]11 /[9]11 . (Then see 5B above).
5D Since x is an element of Z11 , not an integer, it should not appear inside the [ ]11 brackets. If you want
to work that way, start by letting x = [a]11 and then write functions of a inside the brackets.
Rubric (a) 3 marks per operation. Deductions for false statements accompanying the definition.
(b) 10 marks for using familiar algebraic manipulation to get x as a quotient. 4 marks for modular inversion
in Z11 . Deductions: 2 marks per arithmetic error or algebraic slip; larger for more systematic algebra
mistakes.
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