Download proof - Jim Hogan

Proof…
Uniquely Mathematical
and Creative
Jim Hogan, School Support Services
Waikato University
Inspiration from Dr. Paul Brown, Carmel School, WA
Paul presented a similar workshop at MAV Conference, 2008.
I enjoyed the session so much I decided then to present my version
here in Palmerston North. Thanks also to a little book called Q.E.D.
Beauty in Mathematical Proof written by Burkard Polster.
See last slide for all contact details.
What will we learn today?
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Well, who knows? Proof is the pudding!
There are a few problems to ponder
We might experience joy of proof
You might find a useful resource
Your students might benefit
Aspects of the NZC may be made clear
and we must do something with π
Proof
It is nice to experience surprise!
Like a good cryptic clue
P
14. Clothes appear wrong when put on a number (7)
L
Visual Proof
What is the sum of n odd numbers?
1st 2nd 3rd 4th … nth
1 + 3 + 5 + 7 + … (2n-1) = ?
Visual Proof
What is the sum of n even numbers?
1st 2nd 3rd 4th … nth
2 + 4 + 6 + 8 + … 2n = ?
This presents a
nice opportunity to
show the square and
a side;
n2 + n = n(n + 1)
Algebra Re-Vision
A meaning of 1, n and n + 1.
0
1
n-1
n
n+
1
What does 1 look like?
What does n look like?
What does n2 look like?
What does n3 look like?
Visual Proof
What is the sum of n counting numbers?
2 + 4 + 6 + 8 + … 2n = n2 + n
1 + 2 + 3 + 4 +… n = ?
Notice that halving the
even numbers makes the
counting numbers.
How does that help?
Visual Proof
Make the triangular numbers
1st 1
2nd 1 + 2
3rd 1 + 2 + 3
4th 1 + 2 + 3 + 4
and so on…
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 27
406, 425, …
Visual Proof
Two of the same triangular numbers
make n2 + n = n(n + 1)
So one of them is n(n + 1)/2
What is a triangular number?
Visual Proof
What is sum of the multiples of 5?
1st 2nd 3rd 4th … nth
5 + 10 + 15 + 20 + … 5n = ?
Visual Proof
Name other applications
of this knowledge.
See
A Triangular Journey
Powerful Visual Proof
What is sum of the powers of 2?
1st 2nd 3rd 4th 5th … nth
1 + 2 + 4 + 8 + 16 … 2n-1 = ?
Solid Proof
Some of the powers of 2 form a cube.
1 8 64 512 4096 327658
262144
Which ones are they?
An odd look at numbers
nth odd is 2n -1
1 + 3 + 5 + 7 + … (2n-1)
Notice (2n-1) = (n-1) + n.
The odd numbers are consecutive pairs
0+1, 1+2, 2+3, 3+4, 4+5, 5+6, …
Proof by Induction
The idea is easy.
Prove the first is true.
Show for any one k… the truth
Prove for the next one, k + 1… the truth
By a dominoe effect if 1 is true then so is
2 and 3 and 4 and 5 and …so on.
Proof by Induction
Odd numbers are consecutive pairs
0+1, 1+2, 2+3, 3+4, 4+5, 5+6, …
nth odd number = (n – 1) + n
Eg, n = 3, makes 3 – 1 + 3 = 2 + 3 = 5
When n = k, this makes k – 1 + k = 2k - 1
Put n = k + 1, makes k + 1 – 1 + k + 1 = 2k
+1
Notice 2k+1 = 2k-1 +2 is next odd
number. This is INDUCTION.
Prove the sum of the odd
numbers is n2.
When n = 1, n2 is 1
When n = k, n2 is k2
When n = k + 1, k2 is (k+1)2
(k+1)2 = k2 + 2k + 1 =k2 + 2k – 1 +2
We need to see that 2k – 1 +2 is the next
odd number added on… Q.E.D.
Try an Inductive Proof
Sum of whole numbers is n(n+1)/2
Sum of even numbers is n(n+1)
# of Vertices = # of Sides
What happens when you cut off one of the
vertices of a triangle?
What the students may not realise is that
they have performed a proof by induction.
pi time
In a circle of diameter 1, draw a square.
The perimeter πd of the
circle is approximated by the
perimeter of the square
1
2

1
2

1
4
2 2
2
1
2

 1  2.8284...

pi time again
In a circle of diameter 1, draw an octagon.
The perimeter πd of the circle
is approximated by the
perimeter of the octagon
1
2

1
2
1 1
1 1
8
  2    cos 45
4 4
2 2

 4 2 2

 1  3.0614...
The nine digits problem
The digits 1 to 9 have to go into the
nine boxes, with no repeats.
+=
–=
×=
The nine digits problem
Here is one possible solution.
1 + 7 =8
9–4=5
2×3=6
Is there another?
Here are a few solutions
And digits can be swapped between equations.
1+7=8
7+1=8
4+5=9 4+5=9
9–4=5
9–4=5 8–7=1
8–1=7
2×3=6
2×3=6
2×3=6
2×3=6
Why is 2 x 4 = 8 not an option?
The nine digits problem
If three even digits are used in the
multiplication, there are not enough
even digits left for the addition and
the subtraction.
ExE=E
O+O=E
O – O = uh uh
This is called “parity checking”.
That’s it Folks
This file is on my website
http://schools.reap.org.nz/advisor
Clearly labelled for you.There are a
few extra files from Paul and
references to his website and new
book. Thank you…….jim
We could make a conjecture
(Latin “throw together”)
Or could experiment with a supposition
(Latin “place under” & “location”)
Or form an hypothesis
(Greek “under” & “foundation”)
n2 is the sum of the first n odd numbers
n
1
2
3
4
…
k
k+1
2n – 1
1
3
5
7
…
2k - 1
2(k + 1) - 1
n2
1
4
…
k2
??
??
=
k2
+
2(k+ 1) - 1
=
k2
+
2k+ 2 - 1
=
k2
+
2k
=
( k + 1)2
+
1
n2 is the sum of the first n odd numbers
n
1
2
3
4
…
k
2n – 1
1
3
5
7
…
2k - 1
n2
1
4
…
The total of the middle row is 1 + 3 + 5 + … + 2k-5 + 2k-3 + 2k-1
=
1 + 2k-1
+
=
2k
+
=
2k × k/2
3 + 2k-3
2k
=
+
5 + 2k-5
+
…
+
2k
+
…
k2
Proof: the area of a circle is r2
Proof: sum of interior angles of a
triangle is 180
Proof: sum of pentagram
angles is 180
Answers to other “Challenges”
A. For any two numbers, the sum of their squares is never less than twice
their product.
(a – b)2 
0
a2 – 2ab + b2

0
a2 + b2

2ab
Some people like to write Q.E.D. to mark the end of the proof. It is Latin, quod
erat demonstrandum and means “Hooray! I’ve finished the proof!”.
What Q.E.D. does not stand for is “Quite Easily Done”!
Answers to other “Challenges”
B. The sum of any positive number and its reciprocal is 2 or greater.
This is from the excellent
“Proofs without words:
Exercises in visual thinking” by
R. B. Nelson (Mathematical
Association of America)
Answers to other “Challenges”
C. If the final score is a draw, the number of possible half-time scores is a square.
This is from the “Squares” CD-ROM.
Answers to other “Challenges”
D. The product of any two Hilbert numbers is a Hilbert number.
The Hilbert numbers are 1, 5, 9, 13, 17, 21, 25, ...
They are the numbers one greater than numbers in the four times table.
For n = 1, 2, 3, 4, 5, 6, ... the Hilbert numbers are 4n + 1.
Let the two numbers be 4x + 1 and 4y + 1 where x and y are natural
numbers.
The product
= (4x + 1) (4y + 1)
= 16xy + 4x + 4y + 1
= 4(4xy + x + y) + 1
which is a Hilbert number.
Paul Brown can be contacted at
[email protected]