Download Solutions to coursework 3 File

MTH4104
Introduction to Algebra
Solutions 3
Semester B, 2017
Question 1 Let x = 3 + 5t, y = 3 −t and z = 2 − 2t. Calculate xyz +ty · 2x in standard
form a + bt.
Solution This computation could be done in straightforward manner from the definition, but it’s always good to look for shortcuts. Here we spot that
xyz + ty · 2x = xy(z + 2t)
by the distributive law and the commutative law for multiplication. Now
z + 2t = (2 − 2t) + 2t = 2
so we get
xyz + ty · 2x = xy · 2 = 2(3 + 5t)(3 − t) = 2(9 + 15t − 3t − 5t 2 ) =
= 2(4 + 12t) = 8 + 24t.
If you do this the long way using the definition, you have to do more work!
Question 2 Does the pseudo-complex number 3 + t have a multiplicative inverse? If
so, find it; if not, give a proof.
Solution Let’s suppose that the inverse exists and try solving for it. Let a + bt be the
inverse. Then we must solve the equation
(3 + t)(a + bt) = 1
for a and b. Expanding out gives
3a + at + 3bt + bt 2 = (3a + b) + (a + 3b)t = 1 + 0t.
Equating coefficients gives the system of linear equations 3a + b = 1 and a + 3b = 0.
You can solve these in any way you like. For example, multiplying the second
equation by 3 gives 3a + 9b = 0, and subtracting the first equation from this gives
(3a + 9b) − (3a + b) = 0 − 1
i.e. 8b = −1, so b = −1/8. Now substituting this into the second equation gives a −
3/8 = 0, so that a = 3/8. So the inverse sought does in fact exist, and it is
(3 + t)−1 =
1
3 1
− t.
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Question 3 Express the product of i + j + i j and 2 + i − j in standard form α + β j.
Solution In the notation of the preamble to this question, we have α = i, β = 1 + i,
γ = 2 + i and δ = −1. We calculate
αγ − β δ̄ = i · (2 + i) − (1 + i) · −1 = (−1 + 2i) − (1 + i)(−1) = 3i
and similarly,
αδ + β γ̄ = i(−1) + (1 + i)2 + i = −i + (1 + i)(2 − i) = −i + (3 + i) = 3.
So the final answer is 3i + 3 j.
Question 4 Prove that the quaternions satisfy the associative and distributive laws.
(Advice: this is less painful if you represent the quaternions as two complex numbers than if you use four real numbers.)
Solution To do these verifications we simply have to check that given any three
quaternions p = α + β j, q = γ + δ j, and r = ε + ζ j, where α, β , γ, δ , ε, and ζ
are complex numbers, then
(pq)r = p(qr)
and
p(q + r) = pq + pr,
(p + q)r = pr + qr.
Note that since the commutative law for multiplication does not hold in the quaternions,
we cannot treat the two forms of the distributive law as equivalent and must tackle both.
The way these are to be proved is by expanding the sums and products on each side
using the definition, and checking that the results are equal.
I will only do one proof, the proof of associativity; after that you should have no
difficulty (besides attention) doing distributivity yourself. On the left hand side, by the
definition of multiplication,
(pq)r = ((α + β j)(γ + δ j))(ε + ζ j)
= ((αγ − β δ̄ ) + (αδ + β γ̄) j)(ε + ζ j)
= ((αγ − β δ̄ )ε − (αδ + β γ̄)ζ¯ ) + ((αγ − β δ̄ )ζ + (αδ + β γ̄)ε̄) j
= (αγε − β δ̄ ε − αδ ζ¯ − β γ̄ ζ¯ ) + (αγζ − β δ̄ ζ + αδ ε̄ + β γ̄ ε̄) j.
On the right hand side,
p(qr) = (α + β j)((γ + δ j)(ε + ζ j))
= (α + β j)((γε − δ ζ¯ ) + (γζ + δ ε̄) j)
= (α(γε − δ ζ¯ ) − β (γζ + δ ε̄)) + (α(γζ + δ ε̄) + β (γε − δ ζ¯ )) j
= (α(γε − δ ζ¯ ) − β (γ̄ ζ¯ + δ̄ ε)) + (α(γζ + δ ε̄) + β (γ̄ ε̄ − δ̄ ζ )) j
= (αγε − αδ ζ¯ − β γ̄ ζ¯ − β δ̄ ε) + (αγζ + αδ ε̄ + β γ̄ ε̄ − β δ̄ ζ ) j
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where in the second-to-last step we have used the facts z + w = z + w and zw = zw and
z̄ = z for complex numbers to break up the big conjugations. And indeed, the results of
the two expansions are equal.
Question 5 Prove that for any complex number z and quaternion j, the equation
jz = z̄ j
is true.
Solution We have to compute the product of j = 0 + 1 j with the complex number z,
which is z + 0 j when thought of as a quaternion. So setting α = 0, β = 1, γ = z and
δ = 0 and applying the definition we get
αγ − β δ̄ = 0z − 1 × 0 = 0
and
αδ + β γ̄ = 0 × 0 + 1z̄ = z̄.
Hence jz = 0 + z̄ j = z̄ j as required.
Question 6 Let q = α + β j be a quaternion written in terms of complex numbers α
and β . Prove that
qq̄ = |α|2 + |β |2 .
Solution It is possible to do this by applying the formula. But it’s easier to use the
distributive law as follows:
qq̄ =
=
=
=
=
(α + β j)(ᾱ − β j)
α ᾱ + β jᾱ − αβ j − β jβ j
α ᾱ + β ᾱ¯ j − αβ j − β β̄ j2
α ᾱ + β β̄
|α|2 + |β |2
as required. In the third line we used twice the fact from lecture that jz = z̄ j for any
complex number z. In the fourth line we used that ᾱ¯ = α and β α = αβ for any complex
numbers α, β , as well as that j2 = −1.
Question 7 Find, with proof, all quaternions q that satisfy the equation q2 = −1.
Can you describe the set of solutions geometrically?
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Solution For this question I will expand the complex coefficients in the standard form
of q into their own standard form in turn, giving
q = (a + bi) + (c + di) j = a + bi + c j + dk
for real numbers a, b, c, d. This will make the ultimate symmetry of the answer more
apparent. Recall that by definition k = i j.
Using distributivity,
q2 = (a + bi + c j + dk)2 =
= a2+abi+ac j+adk + abi+b2 i2+bci j+bdik + ac j+bc ji+c2 j2+cd jk + adk+bdki+cdk j+d 2 k2 =
= a2+abi+ac j+adk + abi−b2+bck−bd j + ac j−bck−c2+cdi + adk+bd j−cdi−d 2 =
= a2 − b2 − c2 − d 2 + 2abi + 2ac j + 2adk.
This is to equal the real number −1 = −1 + 0i + 0 j + 0k, so equating corresponding
coefficients we have
a2 − b2 − c2 − d 2 = −1,
2ab = 0,
2ac = 0,
2ad = 0.
The last three equations imply that either a or b is zero, and similarly either a or c, and
either a or d. If a 6= 0, then this implies b = c = d = 0, and the first equation reduces to
a2 = −1, which has no solutions when a is real.
So all solutions have a = 0. Substituting this into the first equation gives −b2 − c2 −
2
d = −1, that is b2 + c2 + d 2 = 1. So the set of solutions is
{bi + c j + dk : b, c, d ∈ R, b2 + c2 + d 2 = 1}.
Hopefully you recognise the last equation as the equation for a sphere of radius 1 in
the bcd-space; the answer I was after was “a sphere”. To be precise, the picture corresponding to the Argand diagram for quaternions would be a four-dimensional picture,
since a quaternion has four real components. The solution set lies inside the threedimensional “slice” of this four-dimensional space where a = 0; within this slice it is
the unit sphere.
Question 8 Define a function N : D → R on the pseudo-complex numbers by N(a +
bt) = a2 − b2 .
(a) Prove that, for all pseudo-complex numbers z and w,
N(zw) = N(z) · N(w).
(b) Compute N(1 +t) and N(1). Explain why this computation implies that 1 +t has
no multiplicative inverse.
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Solution (a) The proof of this part is only a matter of expanding. Express z = a + bt
and w = c + dt in standard form. Then we compute that
N(zw) = N((a + bt)(c + dt))
= N((ac + bd) + (ad + bc)t)
= (ac + bd)2 − (ad + bc)2
= a2 c2 − 2abcd + b2 d 2 − (a2 d 2 − 2abcd + b2 c2 )
= a2 c2 + b2 d 2 − a2 d 2 − b2 c2
which is equal to
N(z)N(w) = N(a + bt)N(c + dt)
= (a2 − b2 )(c2 − d 2 )
= a2 c2 − a2 d 2 − b2 c2 + b2 d 2 .
(b) We calculate N(1 + 1t) = 12 − 12 = 0, while N(1) = N(1 + 0t) = 12 − 02 = 1.
If 1 + t had a multiplicative inverse z, so that (1 + t)z = 1, then the result of part (a)
would imply that
N(1 + t) N(z) = N(1) = 12 − 02 = 1.
But N(1 + t) = 0 implies that the left hand side is zero, a contradiction. So 1 + t can
have no multiplicative inverse.
Question 9
(a) Prove that pq = q · p for all quaternions p and q.
(b) Use part (a) to prove that |pq| = |p| · |q| for all quaternions p and q.
(Take extra care not to use the commutative law for multiplication!)
Solution (a) To prove this the only obvious approach is to expand the quaternions
and use the definition of the conjugate. As with some of the later questions, the manipulations will be much less arduous if we write p = α + β j and q = γ + δ j with α,
β , γ, and δ complex numbers, as opposed to the form with four real numbers. Some
facts about how conjugation acts on complex numbers will also be useful: if z and w
are complex numbers, then z + w = z + w and zw = zw and z̄ = z. These were proved in
coursework 2.
By definition the conjugates of our two quaternions are p̄ = ᾱ − β j and q̄ = γ̄ − δ j,
so the product on the right hand side is
q · p = (γ̄ − δ j)(ᾱ − β j)
= (γ̄ ᾱ − (−δ )(−β )) + (γ̄(−β ) + (−δ )ᾱ) j
= (γ̄ ᾱ − δ β̄ ) − (γ̄β + δ α) j
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in which we have used the fact ᾱ = α. On the left hand side, we have
pq = (α + β j)(γ + δ j)
= (αγ − β δ̄ ) + (αδ + β γ̄) j.
We now must take the conjugate of this quantity, which is
pq = (αγ − β δ̄ ) − (αδ + β γ̄) j
= (ᾱ γ̄ − β̄ δ̄ ) − (αδ + β γ̄) j
= (ᾱ γ̄ − β̄ δ ) − (αδ + β γ̄) j
using the facts about conjugation from above. This is equal to what we had before,
proving that pq = q̄ · p̄.
(b) Here the best approach is not to expand the quaternions, but to use the definition
√
|q| = qq̄ which lets us remain in the realm of quaternions.
We would like to prove that |pq| = |p| · |q|, i.e. that
p
√ √
pq · pq = p p̄ qq̄.
It is enough to prove this equation after squaring both sides, i.e. to prove
pq · pq = p p̄ · qq̄.
Using part (a), the left hand side of this is pq · pq = pqq̄ p̄. This is a product of the same
factors as appear on the right hand side, but since quaternions are not commutative
under multiplication, that doesn’t finish the proof. However, the product of the two
middle factors, qq̄, is a real number (by Question 6), and real numbers do commute
with quaternions! So
p · (qq̄) p̄ = p · p̄(qq̄),
and we are done.
Question 10 Suppose we want to define a new number system U to consist of the
numbers of the form a + bu where a and b are real numbers and u is a symbol such
that u2 = 2u − 2. We would like U to satisfy the commutative and associative and
distributive laws.
(a) Provide the formulas we should use to define + and · in U, expressing the sum
and product of two numbers a + bu and c + du in the standard form e + f u.
(b) In fact, U is not really a new number system, but is merely one of the systems we
have already studied in a new guise. Which one? How can you tell?
Solution (a) If U is to satisfy the commutative and associative and distributive laws,
then we can add and multiply in U just the same way we normally would, treating u as
an inert symbol, unless a u2 appears.
As in all the number systems we’ve seen so far, addition is just a matter of adding
corresponding components:
(a + bu) + (c + du) = (a + c) + (b + d)u.
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For multiplication we use distributivity and the fact that u2 = 2u − 2:
(a + bu) · (c + du) = ac + adu + bcu + bdu2 = ac + (ad + bc)u + bd(2u − 2) =
= (ac − 2bd) + (ad + bc + 2bd)u.
Of course, for purposes of writing down a definition of U, we wouldn’t have to show
the intermediate steps; it would be sufficient to write
(a + bu) · (c + du) := (ac − 2bd) + (ad + bc + 2bd)u.
(b) The key insight here is to figure out what u “really is”, in terms of familiar number
systems. What are the solutions to the defining equation u2 = 2u − 2? That equation
can be rewritten
u2 − 2u + 2 = 0
and using the quadratic formula, we find that this has solutions in C, namely
p
√
−(−2) ± (−2)2 − 4 · 1 · 2 2 ± −4
u=
=
= 1 ± i.
2·1
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Since these solutions are in C we might surmise that U is C in disguise. To unmask
it, let’s try to show that if u = 1 + i, then the addition and multiplication formulas are
the same as those in C. (It would work equally well to take u = 1 − i, but we have to
make some choice.)
So if a, b, c, d are real numbers, what is the sum of a + b(1 + i) and c + d(1 + i)? It
is still the case that
a + b(1 + i) + c + d(1 + i) = (a + c) + (b + d)(1 + i),
so our claim for addition holds true. Multiplication is the more interesting case:
(a + b(1 + i)) · (c + d(1 + i)) = ac + ad(1 + i) + bc(1 + i) + bd(1 + i)2
= ac + ad(1 + i) + bc(1 + i) + bd · 2i
= ac + ad(1 + i) + bc(1 + i) + bd · (2(1 + i) − 2)
= (ac − 2bd) + (ad + bc + 2bd)(1 + i).
This also matches our earlier formula for multiplication.
We have thus proven that C and U are two presentations of what is underlyingly the
same number system; you can translate between them by replacing u with 1 + i, or in
the other direction by replacing i with u − 1. (In technical language, for those interested,
what we have shown is that C and U are isomorphic. We will discuss isomorphisms in
lecture more carefully when we come to studying groups.)
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