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. I have used Paul’s presentation and some of his problems , adding some of my own. Proof is a thinking and communicating key competency and as such forms a natural part of the new NZ Curriculum due for implementation in 2010. All students irrespective of ability can enjoy aspects of the enjoyment and experience of proof. This is a critical area of intellectual development necessary for developing understanding rather than living by rules. Enjoy! Nga mihinui Paul!, thankyou.

What will we learn today? 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 We need to identify the learning. In my experience as mathematics advisor this is an area where we need to tread. Students need to be trained to have the expectation of having to think. Progress is sluggish at first. The longer term benefits are huge. In my earlier years of teaching I noticed my students did not “do” as well as similar classes at first. In the last part of the year the same students usually took out some high places and prizes. Perhaps I was teaching for understanding. π

Proof It is nice to experience surprise! Like a good cryptic clue 14. Clothes appear wrong when put on a number (7) P L This is not to set the scene of major mathematics but hopefully some enjoyment, revelation and inspiration. The answer is of course, apparel. The “number is the Roman L or 50. The letters of appear are rearranged. The clue is clothes. Critical, logical, strategic and creative; all blended.

Visual Proof What is the sum of n odd numbers? 1st 2nd 3rd 4th … nth 1 + 3 + 5 + 7 + … (2n-1) = ? The answer is obviously n2. What more does one need? Discuss and record. Let students make this up as it is rich in multiplicative thinking. Use multilink blocks. Discussion is needed to ensure understanding of, for example, the 7 coming from 2x4-1, 7 being the 4th odd number. The 10th odd number is 2x10 -1 = 19 and the first 10 odd numbers sum to 10x10 or 100. Other activities is “make an odd number”. What do two odd numbers make? Make the model! See the square. See the odd numbers. See the sum.

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) Obviously it has something to do with n2 and one side n. Answer is n2 + n = n(n+1). The left hand side of the equation is shown and the right hand side is found by closing the gap and forming a rectangle. Do not rush these ideas and concepts. They are rich in multiplicative thinking. Confirm recognition and understanding of the square and the side. Return to the previous slide as required. Talk and do not rush this. Teachers are new to these ideas.

Algebra Re-Vision A meaning of 1, n and n + 1. What does 1 look like? 1 n n + 1 n - 1 What does 1 look like? What does n look like? What does n2 look like? What does n3 look like? I inserted this slide after trialing the powerpoint. This is essential knowledge of visual proof. Check teachers know this.

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? Answer is n2 + n = n(n+1)/2. We halve the formula! This is not that obvious but we can use another way to confirm. There is a triangular way!

Visual Proof Make the triangular numbers 1st 1 2nd 1 + 2 3rd 1 + 2 + 3 4th 1 + 2 + 3 + 4 and so on… Experience making these wonderful numbers. 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 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? Take time to establish the rectangle n x (n+1). This is the array model of multiplication. Notice dividing by 2 is modeled by breaking the shape into two equal parts.

Visual Proof What is sum of the multiples of 5? 1st 2nd 3rd 4th … nth 5 + 10 + 15 + 20 + … 5n = ? The deduction here is 5 x ½ n(n + 1) and very quickly to the generalisation of the multiples of m. This being mn(n+1)/2. I have asked students to write down all the answers of, say, the 7 times tables 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84 and then as a check have them add them all up. If they could beat me they won a pie at the tuck shop. An irresistible temptation to any teenager. Of course 7 x 78 = 500 + 56 -10 = 546, the answer. Two seconds at the most…I win. Where does 78 come from?

Visual Proof Name other applications of this knowledge. See A Triangular Journey The deduction here is 5 x ½ n(n + 1) and very quickly to the generalisation of the multiples of m. This being mn(n+1)/2. I have asked students to write down all the answers of, say, the 7 times tables 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84 and then as a check have them add them all up. If they could beat me they won a pie at the tuck shop. An irresistible temptation to any teenager. Of course 7 x 78 = 500 + 56 -10 = 546, the answer. Two seconds at the most…I win. Where does 78 come from?

Powerful Visual Proof What is sum of the powers of 2? 1st 2nd 3rd 4th 5th … nth 1 + 2 + 4 + 8 + 16 … 2n-1 = ? This is a lovely and very multiplicative idea. Begin by asking students to model 2, then 2 x 2, then 2 x 2 x 2. The last one will see some students get it all very wrong. This task clearly shows up the “adders” from the “multipliers”. Many students make a big leap forward doing this task. Continue to 2x2x2x2x2x2x2x2 and use this to develop power notation. See next slide also. In this slide notice that 2x the last power, which just happens to be the next power of 2 is one more than the total. Check it out. Hence the sum is 2x2n – 1= 2n+1 -1. Can this be generalised to the powers of m?

Solid Proof Some of the powers of 2 form a cube. 1 8 64 512 4096 327658 262144 Which ones are they? The powers of the form 23n form a cube. Why? Which form squares?

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. Cutting a new side onto a polygon increases the number of vertices and sides by 1. This is a proof by induction. It was true for n=3. It is true for n = k, make k=k+1, and we check it is still true. It is. The number of vertices in a polygon is the same as the number of sides.

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

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

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. E x E = 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 1 2 3 4 … k k + 1 2n – 1 5 7 2k - 1 2(k + 1) - 1 n2 k2 ?? ?? = k2 + 2(k+ 1) - 1 = k2 + 2k+ 2 - 1 = k2 + 2k + 1 = ( k + 1)2

n2 is the sum of the first n odd numbers 1 2 3 4 … k 2n – 1 5 7 2k - 1 n2 The total of the middle row is 1 + 3 + 5 + … + 2k-5 + 2k-3 + 2k-1 = 1 + 2k-1 + 3 + 2k-3 + 5 + 2k-5 + … = 2k + 2k + 2k + … = 2k × k/2 = 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 pbperth@gmail.com