A Mathematics Teacher’s Responsibility –

Beyond the Curriculum

o. Univ. Prof. Dr. Alfred S. PosamentierDean, The School of Education

The City College

The City University of New York

The Extra Responsibilities

• Identifying the problem– Most parents are ill-informed about

mathematics:• They may have had bad experiences with

mathematics.• They may not like mathematics.• They may not know the true purpose of learning

mathematics.• They may not be able to help their children with

mathematics.

The Extra Responsibilities

• Make the parents true partners in the teaching of mathematics– Provide them with the mathematics curriculum– Help them understand the role they are to play at

home to support in-class instruction– Provide them (if necessary) with the information

they will need (e.g. arithmetic algorithms) – Have their children bring home examples of

mathematics that demonstrate its power and beauty

The Extra Responsibilities

• To enrich the curriculum– Acceleration: provide future topics earlier– Expansion: enlarging a topic beyond the

requirement– Extension: move to a related topic beyond

the requirement

– Expose the astonishing in mathematics!

The Most Beautiful Magic Square

The “Melencolia I” by Albrecht Dürer (1471 -1528)

The Magic Square

16 3 2 13

5 10 11 8

9 6 7 12

4 15 14 1

Some properties of this magic square:Sum of all rows, columns and diagonals is 34

16 3 2 13

5 10 11 8

9 6 7 12

4 15 14 1

The four corner numbers have a sum of 34.16 + 13 + 1 + 4 = 34

Each of the four corner 2 by 2 squares has a sum of 34.16+3+5+10 = 342+13+11+8 = 349+ 6+ 4+15 = 347+12+14+1 = 34

The center 2 by 2 square has a sum of 34.10 + 11 + 6 + 7 = 34

The sum of the numbers in the diagonal cells equals the sum of the numbers in the cells not in the diagonals.16+10+7+1+4+6+11+13 = 3+2+8+12+14+15+9+5 = 68.

More properties of this magic square!

16 3 2 13

5 10 11 8

9 6 7 12

4 15 14 1

The sum of the squares of the numbers in the diagonal cells equals the sum of the squares of the numbers not in the diagonal cells.

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

16 +10 +7 +1 +4 +6 +11 +13

= 3 +2 +8 +12 +14 +15 +9 +5 = 748

The sum of the cubes of the numbers in the diagonal cells equals the sum of the cubes of the numbers not in the diagonal cells.

3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3

16 +10 +7 +1 +4 +6 +11 +13

= 3 +2 +8 +12 +14 +15 +9 +5 = 9,248

The sum of the squares of the numbers in the diagonal cells equals the sum of the squares of the numbers in the first and third rows.

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 216 +10 +7 +1 +4 +6 +11 +13 16 3 2 13 9 6 7 12 748

The Fabulous Fibonacci Numbers

Leonardo Pisano

Beginning 1

First 2

Second 3

Third 5

Fourth 8

Fifth 13

Sixth 21

Seventh 34

Eighth 55

Ninth 89

Tenth 144

Eleventh 233Twelfth 377

“A certain man had one pair of rabbits together in a certain enclosed place, and one wishes to know how many are created from the pair in one year when it is the nature of them in a single month to bear another pair, and in the second month those born to bear also. Because the above written pair in the first month bore, you will double it; there will be two pairs in one month. One of these, namely the first, bears in the second month, and thus there are in the second month 3 pairs; of these in one month two are pregnant and in the third month 2 pairs of rabbits are born and thus there are 5 pairs in the month; in this month 3 pairs are pregnant and in the fourth month there are 8 pairs, of which 5 pairs bear another 5 pairs; these are added to the 8 pairs making 13 pairs in the fifth month; these 5 pairs that are born in this month do not mate in this month, but another 8 pairs are pregnant, and thus there are in the sixth month 21 pairs; to these are added the 13 pairs that are born in the seventh month; there will be 34 pairs in this month; to this are added the 21 pairs that are born in the eighth month; there will be 55 pairs in this month; to these are added the 34 pairs that are born in the ninth month; there will be 89 pairs in this month; to these are added again the 55 pairs that are both in the tenth month; there will be 144 pairs in this month; to these are added again the 89 pairs that are born in the eleventh month; there will be 233 pairs in this month. To these are still added the 144 pairs that are born in the last month; there will be 377 pairs and this many pairs are produced from the above-written pair in the mentioned place at the end of one year.

You can indeed see in the margin how we operated, namely that we added the first number to the second, namely the 1 to the 2, and the second to the third and the third to the fourth and the fourth to the fifth, and thus one after another until we added the tenth to the eleventh, namely the 144 to the 233, and we had the above-written sum of rabbits, namely 377 and thus you can in order find it for an unending number of months.”

The Rabbit Problem

a pair of baby (B) rabbits matures in one month to become offspring-producing adults (A), then we can set up the following chart:

Month PairsNo. of Pairs

of Adults (A)

No. of Pairs of

Babies (B)

Total Pairs

Jan. 1 A 1 0 1

Feb. 1 A B 1 1 2

Mar. 1 A B A 2 1 3

Apr. 1A B A A

B3 2 5

May 1 A B A A B A B A 5 3 8

June 1

A B A A B A B A A B A A B 8 5 13

July 1 13 8 21

Aug. 1 21 13 34

Sept. 1

34 21 55

Oct. 1 55 34 89

Nov. 1 89 55 144

Dec. 1 144 89 233

Jan. 1 233 144 377

This problem generates the sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,

The Fibonacci Numbers

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, . . .

Spiral arrangement of the bracts of a pine cone

2 28 5 64 25 39

3 13 39

Find the difference of the squares of the middle two numbers:

Then find the product of the outer two numbers:

2 2

1 1 2n n n nF F F F

Take any four consecutive numbers in the sequence: 3, 5, 8, 13

The Golden Rectangle

w l

l w l

w

l

A

B C

D

2 2

2 2

2

, or

0.

If we let 1, then 1 0.

w wl l

w wl l

l w w

w = 2

51=

2

15 = Φ

1,

The Golden Ratio

11

2

2

1

Then 1 0

1 5Applying the quadratic formula we get:

2

1 51.6180339887498948482045868343656

2

3 2 1 2 1 2 1

4 2 2 1 1 2 2 1 1 2 1 3 2

5 3 2 2 1 1 22 3 1 2 1 3 1 5 3

6 3 3 2 1 2 1 42 4 1 4 1 4 1 8 5

7 4 3 3 2 2 1 6 2 7 2 6 1 7 2 13 8

2

3

4

5

6

7

8

9

10

1 0

1 1

2 1

3 2

5 3

8 5

13 8

21 13

34 21

55 34

The Ratios of Consecutive Fibonacci Numbers

1n

n

F

F

1

n

n

F

F

1

11.000000000 1

1= 1.000000000

2

12.000000000 1

2= 0.500000000

3

21.500000000 2

3= 0.666666667

5

31.666666667 3

50.600000000

8

51.600000000 5

8= 0.625000000

13

81.625000000 8

13= 0.615384615

21

131.615384615 13

21= 0.619047619

34

211.619047619 21

340.617647059

55

341.617647059 34

55= 0.618181818

891.618181818

55

89

55= 0.617977528

144

891.617977528 89

144= 0.618055556

233

1441.618055556 144

233= 0.618025751

377

2331.618025751 233

377= 0.618037135

610

3771.618037135 377

610= 0.618032787

987

6101.618032787 610

9870.618034448

n

n

F

F 1 =n

n

F

F 1 + 1.

There is a single-elimination basketball tournament with 25 teams competing.

How many games must be played in order to get a winner?

Typical Solution:

Any 12 teams vs. any other 12 teams leaves 12 teams in the tournament.

6 winners vs. 6 other winners leaves 6 teams in tournament.

3 winners vs. 3 other winners leaves 3 teams in tournament.

3 winners + 1 team which drew a bye = 4 teams.

2 teams remaining vs. 2 teams remaining leaves 2 teams in tournament

1 team vs. 1 team to get a champion!

Use a chart:

The total number of games played is:

12+6+3+2+1=24

Teams playing Games played Winners

24 12 12

12 6 6

6 3 3

3+ 1 bye=4 2 2

2 1 1

Solution using another point of view:

Consider the losers in the tournament.

There must be 24 losers to get one champion.

Therefore there must be 24 games played

An amazing result:The sum of the squares of the

digits of a given number

You will always end up with either 89 or 1

2 2 2 2 2

2 2 2 2 2

2 2 2 2

2 2

8 9 64 81 145 1 4 5 1 16 25 42

4 2 16 4 20 2 0 4 4 16

1 6 1 36 37 3 7 9 49 58

5 8 25 64 89

2 2 2 2

2 2

52 5 2 25 4 29 2 9 4 81 85

8 5 64 25 89

A (shorter) version that ends up with 1

• This time begin with the number 23:

2 2 2 2

2 2

23 2 3 4 9 13 1 3 1 9 10

1 0 1

Consider the sum of the cubes of the digits of a number

31 1 1 3 3 3153 1 5 3 1 125 27 153 3 3 3370 3 7 0 27 343 0 370 3 3 3371 3 7 1 27 343 1 371 3 3 3407 4 0 7 64 0 343 407

•There are only five numbers that revert back:

•Others will also form a loop – in more steps

The Ulam-Collatz Loop

• Select any arbitrary number

• If the number is odd, then multiply by 3 and add 1

• If the number is even, then divide by 2

• Surprise: You will always end up with 1

An example of the “3n+1” phenomenon

Consider the number: 18 – follow the path:

18 – 9 – 28 – 14 – 7 – 22 – 111 –

34 – 17 – 52 – 26 – 13 – 40 – 20 –

10 – 5 – 16 – 8 – 4 – 2 – 1

Palindromes

RADARREVIVER ROTATOR

LEPERS REPELMADAM I’M ADAM

STEP NOT ON PETSNO LEMONS, NO MELON

DENNIS AND EDNA SINNED ABLE WAS I ERE I SAW ELBA

A MAN, A PLAN, A CANAL, PANAMASUMS ARE NOT SET AS A TEST ON ERASMUS

Palindromic numbers

Here are some palindromic numbers:

121

1331

12345654321

55555555555

To generate palindromic numbers:

• Take any two-digit number and add it to its reversal.

• For example 92 + 29 = 121.

• If you don’t get a palindrome (93 + 39 = 132), then continue the process (132 + 231 = 363).

• Continue till you get a palindrome.

• Caution: 97 requires 6 reversals98 requires 24 reversals!

The Amazing Number 1089 • 1. Choose any three-digit number (where the unit and

hundreds digit are not the same).• We will do it with you here by arbitrarily selecting: 825• 2. Reverse the digits of this number you have selected.• We will continue here by reversing the digits of 825 to get: 528• 3. Subtract the two numbers (the larger minus the smaller)• Our calculated difference is: 825 – 528 = 297• 4. Once again, reverse the digits of this difference.• Reversing the digits of 297 we get the number: 792• 5. Now, add your last two numbers.• We then add the last two numbers to get: 297 + 792 = 1089• Their result should be the same as ours even though their

starting numbers were different from ours. •

If not, then you made a calculation error. Check it.

Let’s look at the first ten multiples of 1089

• Can you see a pattern?

1089 2 21781089 3 32671089 4 43561089 5 54451089 6 65341089 7 76231089 8 87121089 9 9801

1089 1 1089

Getting into an Endless Loop• Choose a 4-digit number (not one with all four

digits the same).

• Rearrange the digits to make the biggest and smallest number.

• Subtract the two numbers.

• With this new number, continue this process.

• Soon you will get 6,174. • But keep going!

• What do you notice?

We will (randomly) select the number 3,203

• The largest number formed with these digits is: 3320.• The smallest number formed with these digits is: 0233.• The difference is: 3087.• The largest number formed with these digits is: 8730.• The smallest number formed with these digits is: 0378.• The difference is: 8352.• The largest number formed with these digits is: 8532.• The smallest number formed with these digits is: 2358.• The difference is: 6174.• The largest number formed with these digits is: 7641.• The smallest number formed with these digits is: 1467.• The difference is: 6174.

• And so the loop is formed, since you keep on getting 6174 if you continue

When is the sum of the digits of a number, taken to a power, equal to

the number?

• Consider these two examples:

2 281 (8 1) 9 81

3 34,913 (4 9 1 3) 17 4,913

Number

= (Sum of the Digits)n

81 = 92

512 = 83

4,913 = 173

5,832 = 183

17,576 = 263

19,683 = 273

2,401 = 74

234,256 = 224

390,625 = 254

614,656 = 284

1,679,616 = 364

17,210,368 = 285

52,521,875 = 355

60,466,176 = 365

205,962,976 = 465

Some Beautiful Relationships

1 1 1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3

4 4 4 4 4 4 4 4 4 4 4 4

1 5 8 12 18 19 63 2 3 9 13 16 20

1 5 8 12 18 19 919 2 3 9 13 16 20

1 5 8 12 18 19 15,057 2 3 9 13 16 20

1 5 8 12 18 19 260,755 2 3 9 13 16 20

1 1 1 1 1 1

2 2 2 2 2 2

1 6 8 15 2 4 9

1 6 8 101 2 4 9

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3

1 5 8 12 26 2 3 10 11

1 5 8 12 234 2 3 10 11

1 5 8 12 2,366 2 3 10 11

More Interesting RelationshipsAmazing!

3 3 3153 = 1 + 5 + 3 4 4 4 41634 = 1 + 6 + 3 + 45 5 5 5 554748 = 5 + 4 + 7 + 4 + 86 6 6 6 6 6548834 = 5 + 4 + 8 + 8 + 3 + 4

7 7 7 7 7 7 71741725 = 1 + 7 + 4 + 1 + 7 + 2 + 5 8 8 8 8 8 8 8 824678050 = 2 + 4 + 6 + 7 + 8 + 0 +5 + 09 9 9 9 9 9 9 9146511208 = 1 + 4 + 6 + 5 + 1 + 1 + 2 + 0

Notice how the powers reflect the original number.

4 6 2 44624 4 4 4 4 1 0 3 31033 8 8 8 8

5 9 5 9 6 8595968 4 4 4 4 4 4

3 9 0 9 5 1 13909511 5 5 5 5 5 5 5

1 3 1 7 7 3 8 813177388 7 7 7 7 7 7 7 7

5 2 1 3 5 6 4 052135640 19 19 19 19 19 19 19 19

Friendly Numbers

• A pair of friendly numbers: 220 and 284.

• The divisors of 220 are: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110.

Their sum is 1+2+4+5+10+11+20+22+44+55+110 = 284.

• The divisors of 284 are:

1, 2, 4, 71, and 142

Their sum is 1+2+4+71+142 = 220.

More pairs of friendly numbers:

1,184 and 1,2102,620 and 2,9245,020 and 5,5646,232 and 6,368

10,744 and 10,8569,363,584 and 9,437,056

111,448,537,712 and 118,853,793,424

The Monty Hall Problem(“Let’s Make a Deal”)

There are two goats and one car behind three closed doors.

You must try to select the car.

You select Door #3

1 2 3

Monty Hall opens one of the doors that you did not select and exposes a goat.

He asks : “Do you still want your first choice door, or do you want to switch to the other closed door”?

1 2 3 Your selection

To help make a decision, Consider an extreme case:

Suppose there were 1000 doors

4 998 9992 99731 1000

You choose door # 1000.

How likely is it that you chose the right door?Very unlikely:

How likely is it that the car is behind one of the other doors: 1-999?

“Very likely”:

1000

1

1000

999

1 3 4 997 998 999 10002

Monty hall now opens all the doors except one (2-999), and shows that each one had a goat.

A “very likely” door is left: Door #1

Which is a better choice?• Door #1000 (“Very unlikely” door)• Door #1 (“Very likely” door.)

4 998 9992 99731 1000

These are all “very likely” doors!

So it is better to switch doors from the initial selection.

Morley’s Theorem for the Angle Trisectors of a Triangle

F

E

D

B

C

A

A Reminder!

It is the teacher’s responsibility to have their students bring home information to excite their parents about mathematics

– so that the parents can then support the importance of mathematics in their child’s education.