The deductive method of proof by Mathematical Induction

1
The deductive method of proof by Mathematical
Induction

• In this session we discuss the method for proving
  the truth of propositions of the form
• ?n P(n)
• where n is any whole number.

2
What is proof ?

• If
• you can convince yourself that the proposition is
  true (or that its negation is false) and
• you can convince a friend (someone who is fairly
  willing to agree with you) that it is true, and
• you can convince (logical) people not willing to
  accept your statements without challenge, that it
  is true,
• then
• you probably have a proof of the proposition.
• Proof is a way of sharing (mathematical) truth.
• Mathematical truths are the consequences of
  assumptions.

3
?n n(n1) 41 is prime.

• Is this proposition true ?
• For n 0,1,2,3,4,5 n(n1) 41 is 41, 43,
  47, 53, 61, 71 which are all prime numbers.
• Is the proposition true for all n ?
• First do we believe it ?

4
List of the 309 primes to 2050

• 2 3 5 7 11 13 17 19 23
  29 31 37 41 43 47 53 59 61
  67 71 73 79 83 89 97 101 103 107
  109 113 127 131 137 139 149 151 157 163
  167 173 179 181 191 193 197 199 211 223
  227 229 233 239 241 251 257 263 269 271
  277 281 283 293 307 311 313 317 331 337
  347 349 353 359 367 373 379 383 389 397
  401 409 419 421 431 433 439 443 449 457
  461 463 467 479 487 491 499 503 509 521
  523 541 547 557 563 569 571 577 587 593
  599 601 607 613 617 619 631 641 643 647
  653 659 661 673 677 683 691 701 709 719
  727 733 739 743 751 757 761 769 773 787
  797 809 811 821 823 827 829 839 853 857
  859 863 877 881 883 887 907 911 919 929
  937 941 947 953 967 971 977 983 991 997
  1009 1013 1019 1021 1031 1033 1039 1049 1051 1061
  1063 1069 1087 1091 1093 1097 1103 1109 1117 1123
  1129 1151 1153 1163 1171 1181 1187 1193 1201 1213
  1217 1223 1229 1231 1237 1249 1259 1277 1279 1283
  1289 1291 1297 1301 1303 1307 1319 1321 1327 1361
  1367 1373 1381 1399 1409 1423 1427 1429 1433 1439
  1447 1451 1453 1459 1471 1481 1483 1487 1489 1493
  1499 1511 1523 1531 1543 1549 1553 1559 1567 1571
  1579 1583 1597 1601 1607 1609 1613 1619 1621 1627
  1637 1657 1663 1667 1669 1693 1697 1699 1709 1721
  1723 1733 1741 1747 1753 1759 1777 1783 1787 1789
  1801 1811 1823 1831 1847 1861 1867 1871 1873 1877
  1879 1889 1901 1907 1913 1931 1933 1949 1951 1973
  1979 1987 1993 1997 1999 2003 2011 2017 2027
  20292039

5
n(n1) 41 for n 0,1,2,,50

• 41 43 47 53 61 71 83 97 113 131 151 173
  197 223 251 281 313 347 383 421 461 503
  547 593 641 691 743 797 853 911 971 1033 1097
  1163 1231 1301 1373 1447 1523 1601 1681 1763
  1847 1933 2021 2111 2203 2297 2393 2491 2591
• cf. the primes
• 2 3 5 7 11 13 17 19 23 29
  31 37 41 43 47 53 59 61 67 71
  73 79 83 89 97 101 103 107 109 113
  127 131 137 139 149 151 157 163 167 173
  179 181 191 193 197 199 211 223 227 229
  233 239 241 251 257 263 269 271 277 281
  283 293 307 311 313 317 331 337 347 349
  353 359 367 373 379 383 389 397 401 409
  419 421 431 433 439 443 449 457 461 463
  467 479 487 491 499 503 509 521 523 541
  547 557 563 569 571 577 587 593 599 601
  607 613 617 619 631 641 643 647 653 659
  661 673 677 683 691 701 709 719 727 733
  739 743 751 757 761 769 773 787 797 809
  811 821 823 827 829 839 853 857 859 863
  877 881 883 887 907 911 919 929 937 941
  947 953 967 971 977 983 991 997 1009 1013
  1019 1021 1031 1033 1039 1049 1051 1061 1063 1069
  1087 1091 1093 1097 1103 1109 1117 1123 1129 1151
  1153 1163 1171 1181 1187 1193 1201 1213 1217 1223
  1229 1231 1237 1249 1259 1277 1279 1283 1289 1291
  1297 1301 1303 1307 1319 1321 1327 1361 1367 1373
  1381 1399 1409 1423 1427 1429 1433 1439 1447 1451
  1453 1459 1471 1481 1483 1487 1489 1493 1499 1511
  1523 1531 1543 1549 1553 1559 1567 1571 1579 1583
  1597 1601 1607 1609 1613 1619 1621 1627 1637 1657
  1663 1667 1669 1693 1697 1699 1709 1721 1723 1733
  1741 1747 1753 1759 1777 1783 1787 1789 1801 1811
  1823 1831 1847 1861 1867 1871 1873 1877 1879 1889
  1901 1907 1913 1931 1933 1949 1951 1973 1979 1987
  1993 1997 1999 2003 2011 2017 2027 2029

6
Proof that the proposition is false

• For n 41
• n(n1) 41 4142 41 41 (421) 41
  43 1763which is not prime.
• For n 40 n(n1) 41 4041 41 412
  1681which is not prime.
• Thus n(n1) 41 is prime for some n, but not
  for all n.
• Proof of existence is complete if you can produce
  one single example, i.e. ? n P(n) is true if
  you can produce an n for which P(n) is true
  e.g.n(n1) 41 is not prime for all n, since
  for n is 89n(n1) 41 is 8397.

7
Proofs for all cannot rest on examples

• Proving that a predicate is true for all
  ?n P(n)
• or that a predicate is true for none, ie. not
  true for any ? (?n P(n))
• cannot be be done by examplesExamples can only
  show that Pr is true, or not true, for some n.
• To prove that a predicate is true for all
  subjects requires a demonstration for any subject
  (independent of the subject).

8
Algebraic deductive proofs

• If some proposition is to be shown ? x, y, etc.
  we take any x, y i.e. we dont give x, y ,
  any values but leave them as unassigned
  variables and follow a chain of implications.
  For example, z (x-y)(xy) ? z x (x y) -
  y (x y)? z x2 x y - y x - y2? z x2 -
  y2Consequently, for any x and y we will
  have (x-y)(xy) x2 - y2

9
Proposition for natural numbers n one, only,
of n, n1, 2n1 is divisible by 3

• 0 1 1
• 1 2 3
• 2 3 5
• 3 4 7
• 4 5 9
• 5 6 11
• 6 7 13
• 7 8 15
• 8 9 17
• 9 10 19
• 10 11 21
• 11 12 23
• 12 13 25
• 13 14 27
• 14 15 29

NB The proposition does not hold for n
0 Do you believe it holds for all other n? Can
you convince a friend? Can you convince anyone
and everyone?
10
Attempted Deductive Proof

• If it does happen to be true for any particular
  set of three numbers, say, m, m 1, 2m 1,
  then, the next set of three numbers will bem 1
  m 2 (2m 1) - (m - 1) if m and m1
  arent, m-1 is2(m1) 1 2m 3 m (m3)
  if m is so is m3
• So if it was the first originally that was
  divisible by 3 in the next set it will be the
  third, if the second then the first, and if
  originally the third, then the second.

11
Proof by Induction

• We know that if for any m one of the triplet m,
  m 1, 2m 1 is divisible by 3, then one of the
  triplet for the next m, i.e. m1 will be
  divisible by 3.
• We know there is a number m0 for which one of the
  triplet m0, m0 1, 2m0 1 is divisible by 3,
  therefore it will be true for all subsequent
  numbers following that m0.
• Recall that we also claimed an a fortiori proof
  earlier.

12

• n 1 2 3 4
  5 6 7 8 9 10 11
  12
• n(1n)(12n) / 6 1 5 14 30 55 91
  140 204 285 385 506 650
• n 2 1
  4 9 16 25 36 49 64 81 100
  121 144
• S n 2 1 5 14 30
  55 91 140 204 285 385 506 650
• Supposing this formula were true, then, it should
  still be true when we add a term. That is, we
  should have

Is it so?
13
Is it so ?
It is !
14
Domino principle

• If any domino falling will knock the next domino
  over
• andany one domino falls
• thenall dominoes after the one that fell will
  also fall over.

15
Proof by Induction

• If
• P(0)1 ie. P is true for n0
• P(m) ? P(m1)
• then
• P(n) is true for all n

16
Binomial Coefficients

• Pascals Triangle
• 1 The
  first column entries are all 1
• 1 2 1 The
  second column entry in row n is n
• 1 3 3 1 The third
  column entry in row n is the
• 1 4 6 4 1
  sum of the first n numbers
• 1 5 10 10 5 1
  These are
• 1 6 15 20 15 6 1 L
  1 3 6 10 15 21 28 36 45 . . .
• 1 7 21 35 35 21 7 1 d(L) 1
  2 3 4 5 6 7 8 9 . . . n
• 1 8 28 56 70 56 28 8 1 mids(d(L)) 1.5 2.5
  3.5 4.5 5.5 6.5 . n1/2
• 1 9 36 84 126 126 84 36 9 1
  S(mids(d(L))) . (1/2) n2 (1/2)n (1)
• 1 10 45 120 210 252 210 120 45 10 1 n(n1)/2
  1 3 6 10 15 21 28 ... Step (1) is a
  sort of pseudo integration.

17
The sum of the first n whole numbers
18
The sum of the first n cubes

• n 0 1 2 3 4 5 6 7
  8 9 10 11 12 13
• n3 0 1 8 27 64 125 216 343 512 729
  1000 1331 1728 2197
• S n3 0 1 9 36 100 225 441 784 1296 2025
  3025 4356 6084 8281
• ?(S n3) 0 1 3 6 10 15 21 28 36
  45 55 66 78 91
• S n 0 1 3 6 10 15 21 28 36
  45 55 66 78 91
• Conjecture the sum of the first n cubes is the
  square of the sum of the first n numbers.

19
A proposition
or
20
Proof by induction
21
Proposition S r 4 f (n)

• where f(n) (1/5)n5 (1/2) n4 (1/3)n3 -
  (1/30)n
• Do you believe it ?
• n 0 1 2 3 4 5 6 7
  8 9
• S n4 0 1 17 98 354 979 2275 4676 8772 15333
• ((n5)/5) ((n4)/2) ((n3)/3) - n/30
• 0 1 17 98 354 979 2275 4676 8772 15333

22
Proof of proposition using the Principle of
Mathematical Induction

• Sn r 4 (1/5)n5 (1/2) n4 (1/3)n3 - (1/30)n
• Is the proposition true for n1 ?
• Can you show that if it is true for nk then it
  is also true for nk1 ?
• If so, you have a proof that the conjecture is
  true for all values of n by the Principle of
  Mathematical Induction.

23
An application to derivatives

• The derivative (rate of change) of a constant
  function is zero, and the rate of change of the
  identity function (yx) is 1.
• If we know the product rule for differentiation,
  then we see that

24
Proposition

• Note first that the proposition is true with m
  1.
• Assume it is true for m n and let's see if we
  can deduce the derivative for m n1

By the Principle of Mathematical Induction the
proposition is true.
25
Whats the big idea?

• The Principle of Mathematical Induction is
  accepted as a way of proving that a statement of
  the form ?n P(n)is true.
• In fact, it is one of Peanos axioms of the
  natural number system.