Download Six more gems that every FP1 teacher should know

the Further Mathematics network
the Further Mathematics network
www.fmnetwork.org.uk
www.fmnetwork.org.uk
Six more gems that every FP1
teacher should know
Let Maths take you Further
General philosophy –
What makes a FP1 gem?
Gem 1: Strong induction
„
For a gem to be precious, it should:
„
„
„
„
motivate students’ interest
aid students’ understanding
reinforce connections within mathematics
„
„
„
Your handout includes notes for each Gem
A game that encourages exploration of
numbers
Shows a surprising result
Encourages a deeper understanding of
mathematical induction
Extends students’ ideas to strong induction
Looking at the problem in a different way
gives insight into mathematical proof
1
Gem 2: Two proofs
„
FP1 students should understand what mathematical
proof is
„
All FP1 students should be aware of these two
classic proofs
‰
‰
„
Euclid’s proof that there is an infinity of prime numbers
Euclid’s proof of the irrationality of root 2, including a
more powerful extension
Could be used in an enrichment session for
potential Further Maths students
2
Gem 3: Rational functions
„
Investigating graphs of rational functions
‰
‰
„
Addresses the misconception that graphs can never cut
their asymptotes
Develops a deeper understanding of rational functions
Extension – ‘The Trouble with Asymptotes’
‰
‰
Cautions against ‘quick methods’
Encourages deeper thinking about limits
Gem 4: Imagine the roots of
a quadratic
„
‘Seeing’ the roots of a quadratic with no real
roots on its graph
‰
‰
Reinforces knowledge of quadratics, particularly
completing the square
Gives some physical reference for complex
numbers
3
Gem 5: Proof by induction
using matrices
„
„
„
Shows students that proof by induction is not
limited to sequences and series
Links matrices and proof by induction
Looks forward to a key FP2 result
Gem 6: Sums of squares
and cubes
„
„
„
A chance to make a model
A physical demonstration of a surprising and
beautiful result
Make sure you always check ‘Maths item of
the month’ on the MEI website!
Two good publications for your
students
„
Encourage your students to look at ‘Plus’
magazine on the internet
„
Take out a departmental subscription to
‘i-squared’ magazine and encourage your
students to subscribe
4
Strong Induction
A game
Think of a number less than 10 (bigger numbers take a long time).
Say you chose 7.
Split into a number pair, say 4, 3. Calculate the product of the pair, 4 × 3 = 12.
Split each member of the pair into a number pair; say 2, 2 and 2, 1. Calculate the
products of the pairs, 2 × 2 = 4 and 2 × 1 = 2.
Split each member of the pairs into a number pair; in this case three new pairs are
possible, 1, 1; 1, 1; 1, 1, because 1 cannot split. Calculate the products of the pairs,
1 × 1 = 1; 1 × 1 = 1; 1 × 1 = 1.
Now there are only 1s left, so no more splitting is possible.
Add together all the totals, 12 + 4 + 2 + 1 + 1 + 1 = 21
This is easier to see on a diagram
Add together the circled numbers, the product of each of
the pairs, to get the total of 21.
Task
•
Use the same process, but split 7 up in different ways.
•
What do you notice?
•
Try different positive integer starting points.
•
What do you notice?
•
Make a conjecture linking the total to the starting number.
•
Prove your conjecture using strong induction. This means that rather than
assuming that the conjecture is true for some particular positive integer, k, you
must assume that it is true for all positive integers from the base case up to and
including the particular positive integer, k.
1
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
Solution
For each positive integer greater than or equal to 2, however you split the integer up
by this process, the total of the products of the pairs is always the same.
Conjecture
If the starting integer is n, the total, g(n) =
1
n ( n − 1) where n ≥ 2 .
2
Proof
g(2) =
1
2 ( 2 − 1) = 1 and 1× 1 = 1 , so it’s true for the base case, n = 2.
2
1
Assume g (r ) = r ( r − 1) for all positive integers 2 ≤ r ≤ k for some positive integer k.
2
We must show g (k + 1) =
1
( k + 1) k
2
Suppose we start with the positive integer k + 1. Choose j such that 1 ≤ j ≤ k and split
the integer k + 1 into two integers, j and (k + 1) – j.
The product of this pair is j ( k + 1 − j )
j ≤ k and k + 1 − j ≤ k
So by the induction hypothesis we have
1
j ( j − 1)
2
1
g (k + 1 − j ) = ( k + 1 − j ) ( k − j )
2
So the sum of the products of the pairs for the
integer k + 1 is
g ( j) =
j ( k + 1 − j ) + g ( j ) + g (k + 1 − j )
= j (k +1− j) +
1
1
j ( j − 1) + ( k + 1 − j )( k − j )
2
2
1
( 2 jk + 2 j − 2 j 2 + j 2 − j + k 2 − kj + k − j − jk + j 2 )
2
1
= (k2 + k )
2
1
= ( k + 1) k
2
=
1
2
So the conjecture is true for n = 2 and if true for all positive integers up to and
including n = k, it is also true for n = k + 1, hence it is true for all positive integers
n≥2.
2
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
Two proofs
All Further Mathematics students should have an appreciation of mathematical proof,
including (hopefully) an aesthetic sense of the beauty of mathematical arguments.
Two classics are
1)
2)
The proof that there are an infinite number of prime numbers (mentioned
as an investigation in the MEI C1/C2 textbook).
The proof that the square root of 2 is irrational (appears in the MEI C3/C4
textbook, but a more powerful extension is discussed here).
In my experience students enjoy discussing these proofs and, through doing so, get a
better understanding of mathematical proof. They can be useful material for a maths
taster session for year 11 students who might be thinking of studying AS Further
Maths.
Students are usually amazed that both of these results were first proved over 2000
years ago.
University mathematics departments report that new undergraduates often have no
appreciation of what mathematical proof is, or of the key distinction between maths
and science; that scientific theories are open to change, or complete rejection, as more
evidence comes to light, whereas mathematical theorems (mathematical results that
have been proved) cannot later be proved wrong unless there was an error in the logic
of the original proof.
Infinity of primes
Euclid’s proof that there is an infinite number of prime numbers - surely every AS
Mathematics student, let alone Further Mathematics student, should be exposed to
this!
Assume that there is a finite number of primes. If so, there would have to be a
greatest prime number, N, which completes the sequence of primes
2, 3, 5, 7, 11, ………. , N
The number Q = 2 × 3 × 5 × 7 × 11 × ........ × N + 1 is not divisible by any of the primes from
2 up to and including N (why?). This means that Q is either prime itself, or is
divisible by a prime greater than N. In either case, a prime must exist that is greater
than N (why?), so there is no largest prime number (why?) and so there must be an
infinite number of primes (why?).
(The first non-prime case is Q = 2 × 3 × 5 × 7 × 11 × 13 + 1 = 30031 = 59 × 509 )
1
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
The irrationalilty of root 2
Euclid’s standard proof by contradiction
If the square root of 2 is rational, it can be expressed as a fraction,
2=
p
, where p and q
q
are integers with no common factors.
2=
p
q
⇒2=
p2
q2
⇒ 2q 2 = p 2
⇒ 2q 2 = ( 2k )
So p 2 is even (why?), so p must be even too (why?)
2
So p = 2k, where k is an integer (why?)
⇒ 2q 2 = 4k 2
⇒ q 2 = 2k 2
So q 2 is even (why?), so q must be even too (why?)
So we have a contradiction (why?), showing that
2
2 must be irrational (why?)
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
A more powerful proof that the square root of 2 is irrational
An alternative and more powerful way of looking at the irrationality of square roots is
to follow Euclid’s argument as far as 2q 2 = p 2 , but them stop at that point because
2q 2 has an odd number of prime factors (why?), whereas p 2 has an even number of
prime factors (why?). Since prime factorisation is unique (hard to prove, but easy to
be intuitively sure of), this is a contradiction (why?). This argument can be extended
to show that the square root of any number that is not a perfect square must be a
prime.
w=
p
q
⇒w=
p2
q2
⇒ wq 2 = p 2
If w is prime, this is a contradiction, so
If w is not prime then …..
w is irrational (why?).
⇒ wq12 q2 2 q32 q4 2 q5 2 ...qn 2 = p12 p2 2 p32 p4 2 p5 2 ... pm 2
Where q1q2 q3 ... and p1 p2 p3 ... are the
prime factors of q and p respectively
⇒ w = pα 2 pβ 2 pγ 2 ... where 1 ≤ α , β , γ ... ≤ m
Why? (given the uniqueness of
prime factorisation)
⇒ w = ( pα pβ pγ ...)
2
So w is a perfect square (why?)
3
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
Investigating graphs of rational functions
Use a graph plotter to explore the graphs of rational functions of the form
(A) y =
ax + b
2
cx + dx + e
(B) y =
ax 2 + bx + c
dx 2 + ex + f
(a) Where possible, find the equations of each type whose graphs have the following
properties:
(i)
Graph has no asymptotes parallel to the y-axis
(ii)
Graph does not cut the x –axis
(iii)
Graph has no asymptotes parallel to the x-axis
(iv)
Graph crosses its asymptote parallel to the x-axis once
(v)
Graph crosses its asymptote parallel to the x-axis twice
(vi)
Graph has an unrestricted range
(vii)
Graph has a range restricted such that l < y ≤ m
(viii) Graph has a range restricted such that l ≥ y or y ≤ m
(b) In each case, use the equation to explain why the graph has this property.
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
ChaDter2: TIIE TROIJBLE WITH ASYMPTOTES
Chapter2: THE TROUBLEWITH ASYMPTOTES
J Deft
Mathetatical Gazette,72, October1988
Considerablestaffroom discussionwas genemtedrecently by a
questionwhich askedfor a sketchof
( i+ 3) ( r + l)
"
( :( - 2)
Clearly there is an asymptotei = 2, and anotheroblique asymptote
whoseequationhasto be found.
"Easy,"saidAloysius."Whenx is very large,the mostsignilicant
term on lhe too line is l. and on rhe boflom is x. The fraclionthen
isJ = t."
becomes(nearinough)f/x, andsothe asymptote
"I don't do it that way," saidBalthazar."Obviously
i+ 4 x+ 3
)=
_-2
.andif we divide the top andbottom lircs by I we get
x + 4 + 3lx
| - 2lr(
'
Whenr is large,3/r and 2lx tendto zerc and we are left with the
asymptote
J = x + 4."
"I don't get either of thoseanswers,"saidCordelia."I agrcewith
*+ 4x+3
'
,- 2
'=
but thenby algebraiclong division I get
)=x+6++.
andwhen.r tendsto infinity this gives anlrr--Oro," , = .x + 6."
As far asthe sketchis concemed,of cou$e, it really doesn'tmatter
-*re graphdisappearsflom the top right-handcomerof the paperat an
angle of approxirnately45"-but clearly only one of theseanswerscan
be dght. Somesimptenumericalwork showsthat the correctsolution
is in facr Codetia's, and it is then fairly easy to seethat each of the
other methodsdependson the erroneousassumptlonthat the limit or
lendency of a quotient is equal to the quotient of the limits,
Nevertheless,the discussioodrew quite forcefully to our aftentionthe
u eliability of tlle "quick methods" that we had be€n using happily
for many yea$.
JOHN DEFT
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page 7
@The MathematicalAssociatiol
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
Chapter2: THE TROUBLE WITH ASYMPTOTES
Questionson Chapter2; THE TROUBLE WITH ASYMPTOTES
1. where doesthe cuve cut ther andt axes?
2. Why is therean asymptoteatt = 2?
3. what happensas'r getscloserto 2 faomaboveard belolv?
4. Illustrate Aloysius'sargumentnumedcallyandexplain how it doet show that ] is very
large andpositive whent is very large andpositive.
5. Explain in the sameway what happensto J when.xis very large andnegative.
6. Sketchthe curve on the basisof fte informationestablishedso far aIId comparevith a
curve plotted on a gaphical calculatoror a graphplotter.
7. Wlat is meantby at oblique asymptote?
8. Show that the line y = t cuts the curve when x = -0.5 and, assumingthe two
branchesof the curve are on opposiiesidesof the asymptote,explain *ith refercnceto
your sketchwhy ) = r cannotbe the obliqueasymptote.
9. Explain Balthazar'satgument.
lO. Showthattheline) = 'x+4cutsthecu
be the oblique asymptote.
I L Explainhow cordeliashowedthat{j3=
ewhen] = -5.5 andexplainwhy it cannot
= x + 6 +
.--ll.
12. Why doesthis showthatl, = x+ 6 is the obliqueasymptote?
13. What happensif you try to find wherc) = .x + 6 cuts the curve?
14. What doesit mefunbythe erroneousassunption that the limit of a quotient is equal to
the quotientof the limits?
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page8
O The Maftematical Association
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
Answe$ to ChaDter2: TIIE TROUBLE WITH ASYMPTOTES
8. ) = x meetsthe curvewhen.x =
(.x+3)(x+1)
(x-z)
-x (xf-2-)2=x( r=+*3+)4(xr +. +13)
=
=
-6x=3
=
x = -0.5
if we assume
thetwo branches
This meansthat/ = r cannotbetheobliqueasymptote
assumption?)
of thecurvelie on oppositesidesof theasymptote.(Is this a rcasonable
t'
3t* 3. n""uur"
fractionfor ), i.e. J =
9. Balthazarhaslookedat the expanded
I
he doesl't want r in the denominatorgettinglarger,he dividei boir pans of the
fractionby r.
v+a +J
1
)
':----:-=---1.
Now I and tr both becomevery small as r increases,so as t
So y =
r -:
x
t
becomesvery large, it appearsthat both of thesecan be ignored,reducingthe fraction
lhe asymplote
is y = y + 4.
to
This suggests
-.
x2 +4 x+3
*+4x+3
1 0 . y = 'r* 4c uts y=
wherex + 4 =
w -)
-+
(x + 4 ) ( x - 2 )
x 2+ 2 x - 8
_t-
f+4x+3
*+4x+3
ll
This showsthat it cannotbe an obliqueasymptotefor the sarnereasonasin question8.
1,1. Cordelia
x2 + 4x + 3 canbe divided by,r - 2 asfollows;
x+6
x -2
f +4 x+ 3
*-2 t
6.r+3
So .x2 +4 .r+3 =
and
.f+4x+3
/* 1 4 \/--r\rl <
=,r+o +---- = .
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
pqge10
O The MathematicalAssociation
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
Answelsto Chapter2: THE TROUBLEWITH ASYMPTOTES
very small,until its sizeis negligibleandthe
verylarge,
becomes
12. As .r becomes
---iivalueofj|, tendsto, + 6, i.e,y = r + 6 is theasymptote.
i +4x + 3
t5. y = x + ocutsJ= --;:--wnerer
.
l+ 4 x + 3
+o =
+(r+ 6 ) ( . x - 2 ) = f + 4 x + 3
-=T= *+4x+3
=
*+4x-12
-12=3
which is nonsens€,
i.e. we cannotfind a value for x which fits this. This means
i +4x +3
=
y=
x_ 2
n e ve ctosses]
r
J+ o.
14. Both Aloysius and Balthazar had looked at the limits of the numemtor and the
denominato.separatelyasx becamevery large and haddivided theselimits to look for
drc asymptote,i.e. they had investigatedthe "quotient of the limits". They had
assumedthat this would give the limit of the expressionfor X, i.e. "the limit of the
ouotie[t".
Cordelias work showedthat thesemethodsdid not give the corect answer,showing
that Aloysius and Balthazarhad b€€n wrong to make this assumption,i.e. it rvas alr
"eroneous assumption".
In a similar questionit would be w'ong to look at the denominatorand numentor
separatelyto extractinformation abouttheir effect togedrcr.
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page11
@The MathematicalAssociation
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
Chapter6a: IMAGINE THE ROOTSOF A QUADRATIC
Chapter 6a: IMAGINE THE ROOTS OF A QUADRATIC
C R Holmes
MathematicalGazette.74. October1990
At school, I was introducedto complex numbersby consideringa parabola
drifting upwards:
Here it has
two rcal distinctroots.
Here it has
two rcal equalroots.
Here it has
no real roots.
We were eventually shown that all quadraticshave two roots, and shown
how to find them. But I wantedto seethem on the graph! It was explainedto
me that you can't show them on the graph-{hat if the paraboladoesn'tcut the
axis then it's just hard luck-you can calculatethe roots, but not showthem on
the gaph. This was most unsatisfying:I felt sure there should be some
relationship
belweenthe rootsandthediagram.
Now, manyyearslater,I havefounda solution.I cannow "see"therootsof
a quadraticequation,given its gaph-and I cando this instantlyandeasily!
We can still use the idea of a paraboladrifting upwards.But, this time,
imaginetheparaboladlagswith it a reflectionof itself,like this:
The reflection is tied to the "nose" of the original parabola, so that, as the
original parabolarises,so doesthe reflection.
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page26
O The MathematicalAssociation
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
ChaDter6a: MAGINE THE ROOTSOF A QUADRATIC
We canusethe reflection to "see" the roots,when the original parabolahas
drifted abovethe x-axis.
y=l-ax+s
hassolutioN 3 t 2.
(ii)
t=f
-6x +9
(i n)
J =f
-5;c +13
hassolutions3 t 0.
hassolutions3l 2i.
In genemlby imagining a reflection of the parabolain this way one cansee
its imaginaryloots: the interestedreadercaneasily checkthis'
C R HOLMES
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page27
Association
O TheMathematical
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
ChaDter6b: THE COMPLEX ROoT OF A QUADRATIC FROM ITS GRAPH
Chapter 6b: THE COMPLEX ROOT OF A QUADRATIC
FROM ITS GRAPH
S F G Wessels
Gazette,65' March1981
Mathematical
Let the complexroots of the quadraticequationf (r) = I + px + q = o
beb t ic. Then
f(x)=(r - b + ic)(x - b - ic)
=x2 -zb x+ b2+c2
=(x-b )2 +c2,
Thereforeb is the .r value wherethe parabolay = f(t) hasa minimum value
Draw thegraph, = f(r). Then(b, c2)is theminimumpoint ofthe curve.
The line y = 262qutsy = f (x) in thepointsC (, - c, 2c2)andD (b + c,
lc') slnce
f ( . r ) = ( . t - b f + & = zcz
:. (x-D2 =C
= b - c.
-'. x = b + co r|
From the graphvr'ecanfind c2andwe canthusdraw the line y = 2c2
F-
c --------t-
c------,1
From the graphwe cannow seethe valuesof b and c.
S F G WESSELS
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
pdge 28
O The MathematicalAssociation
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
A
Questionson Chapter6a: IMACINE THE ROOTSOF QUADRATIC
l. (a) By factorising,confim the solutionsin cases(i) and (ii)'
(b) How doesthis form provide the roots?
2. If an equationhasreal roots, explain why the solutionsar-e of the form m t r,' where
.:r = mls the line of symmetryof the conesponding$aph'
3. By using the quadraticformula, showcase(iii) hasno real solutions'
4. (a) complete the squarein case(iii).
(b) How canyou deducefrom this form that thele are no real roots?
(c) How can the complexroots be identified from this form?
in
5. Use the completedsquareform ofr2 - 6' + 13 to explain why the rcflected curve
= I - (x - 3)2 Use these
case (iii) has an €quatron which can be writien i
to explain the reflection
curves
tqu-" forms for the equationsof the two
"oapf"t"a
methodfor finding imaginaryroots.
6. The generalform for a quadraticcan be written y = f + bt + c -Reanangethis by
= 0' What
the square.Rewrite to give an exprcssion for 'r when )
"o-p-t"ting must b andc fulfil for thereto be real roots?
conditions
Quesrionson chapter 6b:
THE COMPLEX ROOT OF A QUADRATIC
FROM ITS GRAPH
equation/ (.t) .=,(x.- b\2 +. c2 given in this article'
?. By rcferenceto the general
-y = 8 all-owsus to find
the complex roots of the equation
why the line
;phi.
x2 - 6r + 13 = 0 in question6.
8. Use a similar methodto the oneillustated in the anicle to fitld the complexroots of:
(a 't f
h\
+6 x+2 5 =O
L 1 t2 +3 x+2 =0
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page29
@The MathematicalAssociation
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
Answersto Chapter6a: IMAGINE THE ROOTSOF A QUADRATIC
Answersto Chapt€r 6a: IMAGINE THE ROOTS OF A QUADRATIC
l.
( a ) C ase(i )
Case(ii)
x2 -6 x+5 = ( .{ - 5) ( ,r - l) ther cforreootsatr = 5andt = 1.
.r'?- 6x+9 = (x- 3)(x- 3) thereforedoubleroot at t = 3.
(b) By equatingeach brucketto zero, as their product must equal zero where roots
occur.
2, This usesthe symmehicalFoperties of the parabola.
plac€don eithersideofx = m.
The rootsarcsymmetrically
(Also considerhow completingthe squareor using the quadratic'formula' works.)
3. * + fu + c = 0 hasno realrootsif 12 - 4ac < 0.
I n l - 6.r+ 1 3 = 0 , a = l , b =4 andc= 13,so b2- 4ac= - 16<0. Tler efor etle
equationhasno realrcots.
(x - 3) 2+4.
4 . ( a ) 1 , = f -A t+ 1 3 +y=
(b) By completingthe squareit canbe seenthat the minimum value of f is 4. This tells
us that the curyenevercrossesthe x-axisand so the equationx2 - 6r + 13 = 0
hasno real roots.
(c) ("r- 3)'?+ 4 = 0 + x - 3 ='i/4 = !2i + x = 3 r 2i.
5 . P - e x + 1 3 = (,x-3 f +4 .
when the curve is reflected in the line ) = 4, the equation changes from
f = 4+ (x - 3)2 to ) = 4- (x- 3)2. The roots cofiespondingto the equations
4 + ( x - 3 \x = 0 a n d 4 - (x - 3 )2= 0ar ethen3!2iand3 !2.
6 . ! = x 2 + b x +c.
Completingthe square,t =
I
h\'
*
*,
t"
;)
-
b2
i
=
y=u^ t b+\2 b 2
f.Y*t) i-r.
-bt
'
=
glves,r
r<ear:mgrng
G
2
--!!,ro6'
- 4c ) 0 for Lhere
ro berealroors.
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page30
@The MathematicalAssociation
From 'Pig and Other Tales' by Doug French and Charlie Stripp, published by the Mathematical Association
Answe6 to Chapter6b: THE COMPLEX ROOT OF A QUADRATIC FROM ITS CRAPH
Answersto Chapter6b: THE COMPLEXROOTOF A QUADRATIC
FROM ITS GRAPH
1. Wtr'ry = f -6x+ 13 is wr:ittenasa completedsquareit becomesy = (l - 3.12+ 4
The + 4 in this caserepresentsthe +c2, so *re line y = 2c2 becomesthe line y = 3,
which can,by comparingwith the other method,be thoughtof asreflection of the r-axis
i.;l = +.
8. (a) y = f +6x+ 25, completethe square:y= 1x+3)2+ 16, so the roots of
Q + 3f + 16 = 0 are symmetrical about .{ = -3 and the minimum point is
(-3,16). The rootsof thee4uationarefoundby equatingc2with 16to givec = 14,
so givingr = -3 + 4i andx = -3 - 4i,
(b) y = u2 + 3t + 2, divide though by 2: | = S a lx + 1, completethe square:
yt =
1x + ]\2 + *, so the roots are symmetricalabout r = f and the minimum
point is 1f, rt). The roots of the equationare found by equatingC with rt to give
c = t * , s o . r = - ] + i { o r x = + - i*.
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
page31
@The MathematicalAssociation
Proof by induction with matrices
This is not directly on the FP1 syllabus, but it’s a nice application of proof by
induction that will reinforce students’ skills and understanding of both proof by
induction and matrices.
It’s also important for students to understand that proof by induction has applications
outside series and sequences.
Question (taken from the Pure 4 examination, January 2005)
⎛ −2 9 ⎞
⎛ 1 − 3n
n
Given that M = ⎜
⎟ , prove by induction that M = ⎜
−
1
4
⎝
⎠
⎝ −n
positive integer.
9n ⎞
⎟ , where n is a
1 + 3n ⎠
Solution
⎛1 − 3 × 1
When n = 1, M n = ⎜
⎝
−1
9 × 1 ⎞ ⎛ −2 9 ⎞
⎟=⎜
⎟ = M , so conjecture true for n = 1.
1 + 3 ×1⎠ ⎝ −1 4 ⎠
⎛1 − 3k
Assume true for n = k, so M k = ⎜
⎝ −k
9k ⎞
⎟
1 + 3k ⎠
9k ⎞
⎛ −2 9 ⎞ ⎛1 − 3k
⎟⎜
⎟
⎝ −1 4 ⎠⎝ −k 1 + 3k ⎠
⇒ M k +1 = MM k = ⎜
⎛ −2 + 6 k − 9 k
−18k + 9 + 27k ⎞ ⎛ −2 − 3k 9k + 9 ⎞
⎟=⎜
⎟
⎝ −1 + 3k − 4k −9k + 4 + 12k ⎠ ⎝ − k − 1 3k + 4 ⎠
⎛ −1 − 3 ( k + 1)
9 ( k + 1) ⎞
=⎜
⎟
⎜ − ( k + 1)
1 + 3 ( k + 1) ⎟⎠
⎝
=⎜
But this is the conjecture with k + 1 in place of n, so if the conjecture is true for n = k,
it is also true for n = k + 1, and since it is true for n = 1, it is true for all positive
integers n.
Extension
In FP2 students learn that, for a square matrix, M n = SΛ nS-1 , where S is the matrix of
the Eigen vectors of M and Λ is the matrix of the Eigen values of M, so that MS =
SΛ. At FP1, you could give students the result M n = SΛ nS-1 , together with the
equation MS = SΛ, and challenge them to use their knowledge of matrix algebra to
prove the result by induction.
This reinforces both proof by induction and matrix algebra skills and for students who
go on to FP2, it gives them a taste of what’s to come.
1
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
Question
Prove by induction that M n = SΛ nS -1
Solution
When n = 1,
M = SΛ S -1 = MSS −1 = M (using MS = SΛ), so conjecture true for n = 1.
Assume true for n = k, so M k = SΛ k S -1
⇒ M k +1 = MM k = MSΛ k S -1 = SΛΛ k S -1 = SΛ k +1S -1 (using MS = SΛ)
But this is the conjecture with k + 1 in place of n, so if the conjecture is true for n = k,
it is also true for n = k + 1, and since it is true for n = 1, it is true for all positive
integers n.
2
‘Six gems that every FP1 teacher should know’, MEI Conference 2008
Charlie Stripp
⎛ n ⎞
Showing that ∑ r = ⎜ ∑ r ⎟
r =1
⎝ r =1 ⎠
n
2
3
n
1. Print off the two pages below and stick them on to separate sheets of
sturdy card. The card used on a typical pad of A4 paper works well.
n
∑r×r = ∑r
2
r =1
3
r =1
2. Using the perimeter outlines, cut out these two pieces of card. For each
shape, cut along the diagonal and then take the piece above the diagonal
and make the two vertical cuts down to diagonal so that you are left with
four shapes from each piece: the big piece and three trapezia.
Place paper-side down on the table.
4. Stick the two large pieces of card back-to-back and leave to set.
5. Insert the three pins into the corresponding point in the large shape
between the two pieces of card. If this is done correctly it should be
possible to rotate the trapezia so that the top shape (on the right) is
transformed into the bottom shape.
⎛ n ⎞
⎜∑r⎟
⎝ r =1 ⎠
2
2
⎛ 4 ⎞
This demonstrates ∑ r = ⎜ ∑ r ⎟ . This could be done numerically –
r =1
⎝ r =1 ⎠
what do you need to show to turn this demonstration into a general proof?
4
3
1+2+3+4
© MEI 2007
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
4×42
3×32
2×22
1×12
3. Place one pin as shown on one trapezium so that it perpendicularly bisects
the diagonal edge. Using wood glue or other strong glue, stick together the
two trapezia of the same size so that the pin is between the card and the
paper is outermost. Repeat for the other two pairs of trapezia.
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
© MEI 2007
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp
© MEI 2007
The finished article
'Six gems that every FP1 teacher should know', MEI Conference 2008, Charlie Stripp