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Year 9 Proof Dr J Frost ([email protected]) Objectives: Understand what is meant by a proof, and examples of what does and what doesnβt constitute a proof. Be able to solve problems involving sequences, consecutive numbers and involving digits. Last modified: 19th February 2015 How Many Examples Needed? In 1772 Euler noticed that the following equation gives prime numbers for many positive integers n: ππ β π + ππ We might wonder if the statement βFor all positive integers n, n2 β n + 41 is primeβ is true. Try this for a few π. How many examples of n would we need to show this statement is: a) True One ο» A few ο» Infinitely οΌ many b) False One οΌ A few ο» Infinitely ο»many The smallest value of π for which this statement is false is: 41? Counterexamples ! A counterexample is an example used to disprove a statement. Discussing in pairs, find counterexamples for the following statements: 1 Statement Possible counterexample Prime numbers are always odd. 2 is prime, but not odd. The square root of a number is 2 always smaller than the number itself. 3 If π is prime then π + 2 is prime. 2π2 + 11 is prime for all 4 integer values of π ? 0.25 = 0.5, but 0.5 > 0.25. ? 7 is prime but 9 is not. ? (2 × 112 ) + 11 will be divisible by 11 ? because 2 × 2 11 and 11 both are. SKILL #1: Sequence Proofs 1st 2nd 3rd 4th 5th 5 4 9 13 22 35 57 2 6 8 14 22 36 58 Each of the numbers is the sum of the two numbers to the left. You get to set the first 2 numbers (indicated in black). a) Try a few starting numbers, can you discover a rule that allows you to quickly work out the last (5th) number given the first two, without having to work out the ones in between? b) In this example we had 5 squares after the first two. Can you generalise your rule to work for any number of squares? Solution ! To represent something generically, use an algebraic variable. a ? ? ? ? b a+b a+2b 2a+3b 3a+5b 5a +? 8b Our rule could be this: The last number is 5 times the first one plus 8 times the second one. Generify! For the nth square, we multiply the first number by the nth Fibonacci ? number and the second number by the (n+1)th Fibonacci ? number, and add the results. Exercise 1 Q1 Q2 Find a rule which gives you the orange number given the black numbers(s). 6 5 a b 1 4 -3 7 -10 a-b 2b-a ? 2a-3b 5b-3a Q3 5a-8b [Cayley] The first two terms of a sequence are the numbers 1, 2. From then on, each term is obtained by dividing the previous term by the term before that. Thus the third term is obtained by dividing the second term, 2, by the first term, 1. (a) Write down the first five terms. ? π, π, π, π, (b) π π Calculate the fiftieth term. π π Terms repeat every six: π, π, π, π, π , π , π, π, β¦ π ? π If 6th term is π, then 48th term is π, and so 50th term is π. (c) What happens if other non-zero numbers are chosen for the first two terms, but the rule for calculating the next term remains the same? (Not, you must prove this behaviour in general!) If first two numbers are π, π. Then subsequent numbers are π π π π , , , , π, π, β¦ π π π π Since each term is based only on the previous ? two, we have proved the sequence always cycles every 6 numbers for any two starting numbers. N The first and second terms of a sequence are added to make the third term. Adjacent odd-numbered terms are added to make the next even-numbered term, for example, First term + third term = fourth term And third term + fifth term = sixth term Likewise, adjacent even-numbered terms are added to make the next odd-numbered term, for example, Second term + fourth term = fifth term Given that the seventh term equals the eighth term, what is the value of the sixth term? Terms are π, π, π + π, ππ + π, ππ + ππ, ππ + ππ, ππ + ππ, ππ + ππ Since 7th term = 8th term, ππ + ππ = ππ + ππ. Therefore π = βπ Thus sixth term is ππ + ππ = βππ + ππ = π [Hamilton] The first two terms of a sequence are the numbers 1, 2. From then on, each term is obtained by adding 1 to the previous term and then dividing by the term before that. Thus the third term is obtained by adding 1 to the second term and then dividing by the first term. a. Write down the first five terms. π, π, π, π, π b. Calculate the sixtieth term. Continuing, 6th term is 1 and 7th is 2. So sequence repeats every 5 terms. If 5th term is 1, the 60th term is also 1. c. What happens if the other non-zero numbers are chosen for the first two terms, but the rule for calculating the next term remains the same? ? ? ? If first two terms are π and π, then we get π+π π+π+π π+π ? π, π, π , ππ , π , π, π. This requires some fairly difficult algebraic simplification! Reasoning about sides and angles. Introducing variables for unknowns. Algebraic Geometric Some types of proofs Games If I have a βperfect Connect 4β strategy for winning, how can I show itβs perfect? Even/odd e.g. βShow that n2 + n is always even.β SKILL #2: Divisibility Proofs Show that the sum of any three consecutive integers is a multiple of 3. Kyleβs proof: β5 + 6 + 7 = 18, which is divisible by 3β. We could represent 3 consecutive integers as: π π+π π+π ? Then: π+ π+π + π+π = ππ + π = π(π +?π) which is divisible by 3. Que Problemo? He hasnβt shown itβs true for all? possible integers. SKILL #2: Divisibility Proofs How could I algebraically represent: ! An odd number 2π?+ 1 An even number 2π ? Two consecutive odd numbers 2π + 1, ? 2π + 3 (but 2π β 1 and 2π + 1 might make your subsequent algebra easier) The sum of the squares of two consecutive numbers. A power of 2. One less than a multiple of 5. π2 + π + 1 ? 2?π 5π?β 1 2 Check Your Understanding Prove that the sum of three consecutive even numbers is a multiple of 6. 2π + 2π + 2 + 2π + 4 = 6π + 6 = 6(π + 1) ? which is a multiple of 6. Prove that the product of two consecutive odd numbers is always one less than a multiple of 4. 2π + 3 2π + 1 = 4π2 + 8π + 3 = 4 π2 + 2π + 3 Which is 3 more than a multiple of 4, i.e. 1 less than a multiple of 4. Alternatively: ? 2π + 1 2π β 1 = 4π2 β 1 Which is clearly one less than a multiple of 4. Exercise 2 Q1 Prove algebraically that the sum of two consecutive odd numbers is divisible by 4. ππ + π + ππ + π = ππ + π = π(π + π) N1 ? Q2 [GCSE] I think of two consecutive integers. Prove that the difference of the squares of these integers is equal to the sum of the two integers. Two numbers are: π and π + π Difference of squares: π + π π β ππ = ππ + π Sum of numbers: π + π + π = ππ + π These are equal. ? Q3 π + π π β ππ = ππ + πππ + ππ + π β ππ = πππ + ππ + π = ππ π + π + π The product of two consecutive integers is even, thus ππ(π + π) is divisible by 6. ? N2 Prove that the difference between the squares of two odd numbers is a multiple of 8. ππ + π π β ππ β π π = πππ + ππ + π β πππ β ππ + π = πππ + ππ + π β πππ + ππ β π = ππ which is divisible by 8. [JMO] Find a rule which predicts exactly when five consecutive integers have sum divisible by 15. π+ π+π + π+π + π+π + π+π = ππ + ππ = π(π + π) Thus to be divisible by 15, π + π must be divisible by 3, i.e. the middle number of the five numbers must be divisible by 3. ? [IMO] Prove that there is exactly one sequence of five consecutive positive integers in which the sum of the squares of the first three integers is equal to the sum of the squares of the other two integers. π β π π + ππ + π + π π = π+π π+ π+π π πππ + π = πππ + πππ + ππ ππ β πππ β ππ = π π + π π β ππ = π π = βπ ππ π = ππ So ππ + πππ + πππ = πππ + πππ is only solution. ? ? Q4 Prove that the difference between two consecutive cubes is one more than a multiple of 6. N3 Prove that the product of four consecutive numbers is one less than a square number. π π+π π+π π+π = ππ + π ππ + ππ + π = ππ + πππ + ππππ + ππ + π ? = ππ + ππ + π π SKILL #3: Digit Proofs βI think of a two-digit number. I then reverse the digits. Prove that the difference between the two numbers is a multiple of 9β e.g. 71 β 17 = 54 Let βabβ be a two digit number. Then its reverse is βbaβ. The value of each number is 10πAlgebraically + π and 10π +?π Model Problem respectively. Thus their difference is 10π + π β 10π + π = 9π β 9π β π)conclusion (Note how factoring Manipulate =to9(π reach ? out the 9 is a good explicit way to show the which is a multiple of 9. expression is a multiple of 9) Test Your Understanding a What would be the value of the 4-digit number with digits "ππππβ? πππππ + ππππ + πππ + π b What would be the value if its digits were reversed? πππππ + ππππ ? + πππ + π ? c What is the sum of these two numbers? πππππ + ππππ?+ ππππ + πππππ d Hence prove that the sum of a four digit number and its reverse is a multiple of 11. = ππ(πππ + πππ ? + πππ + πππ ) Rubbish Proofs Darth Kittyβs solution: βPick the number 31 for example. The reverse of this is 13. And 31 β 13 = 18, which is divisible by 9. Hence the statement is true.β What is wrong with their proof? Showing something is true for one example is not sufficient. We need to show the ? POSSIBLE starting statement is true for ANY number we choose. Rubbish Proofs Photoshop Kittyβs solution: βHereβs all possible two-digit numbers: 11 β 11 = 0, 12 β 12 = 0, ... 21 β 12 = 9, 22 β 22 = 0, 32 β 23 = 9, 42 β 24 = 18, ... 98 β 89 = 9, 99 β 99 = 0 These are all divisible by 9. What is wrong with their proof? Technically itβs a valid proof, because weβve shown the statement is true for every possible two-digit number. But itβs bad maths having to list out every?case, and would become difficult if for example we tried to generalise the statement to 3-digit numbers. Proof of 9 divisibility rule Prove for 2-digit numbers that if the sum of the digits of a number is a multiple of 9, then the number itself is divisible by 9. How to represent a two digit number π with digits "ππ": π = πππ + π ? How do we represent property that digits add up to a multiple of 9? π + π = ππ ? How do we combine these equations to therefore show that the original number π is a multiple of 9? π = πππ + π = ππ + π + π = ππ +?ππ =π π+π Hence π is divisible by 9. Exercise 3 β Digit Problems [JMO] Observe that 49 = 4 x 9 + 4 + 9 I think of a 3-digit number and then reverse its digits. Q1 Prove that the difference between these two numbers 4 Use algebra to find all other two-digit numbers which are equal to the product of their digits plus the sum of their digits. is a multiple of 11. If a number has the digits βπππβ then its reverse is "πππβ. Each number has the value ππππ + πππ + π and ππππ + πππ + π. ππππ + πππ + π β ππππ + πππ + π = πππ β πππ = ππ ππ β ππ πππ + π = ππ + π + π ππ = ππ Given π is not 0: π=π Thus numbers are ππ, ππ, β¦ , ππ ? [IMO] All the digits of a certain positive three-digit Q2 number are non-zero. When the digits are taken in ? Prove that there are no three-digit numbers b which are equal to the product of their digits plus the sum of their digits. reverse order a different number is formed. The difference between the two numbers is divisible by eight. Given that the original number is a square number, find its possible values. Difference of βπππβ and "πππβ is ππ(π β π) Since 99 is not divisible by 8, π β π is. Only possible if π = π, π = π or vice versa. Only square in former case is 169. Only square in latter case is 961. ? [IMO] An βunfortunateβ number is a positive integer Q3 which is equal to 13 times the sum of its digits. Find all βunfortunateβ numbers. (Hint: start with 2 digit first) For two-digit πππ + π = ππ π + π β ππ + πππ = π which is impossible. Using same approach for 3digit: 117, 156, 195 Need to also show not possible for more than 3 digits. ? ππππ + πππ + π = πππ + π + π + π πππ + ππ = πππ But since ππ is at most 81, ππ > ππ and thus πππ > πππ and hence the LHS is always greater. ? 5 N Can you prove the 9 divisibility rule for 3 digit numbers? A divisibility test for 7 is to subtract twice the last digit from the remaining numbers, and see if the result is divisible by 7. e.g. 392 -> 39 β 4 = 35, which is clearly divisible by 7, so 392 is. Prove this works for all threedigit numbers. Let our number be n with the digits βabcβ. π = ππππ + πππ + π. If we subtract twice the last digit from the remaining numbers, we get 10a + b β 2c. Now if this is divisible by 7, 10a + b β 2c = 7k for some integer k. Thus ππππ + πππ β πππ = πππ Then n = 100a + 10b + c = 70k + 21c = 7(10k + 3c) which is divisible by 7. ? Another Example of Good/Bad Proofs [IMO] βAn arithmetic sequence is one in which the difference between successive terms remains constant (for example, 4, 7, 10, 13, β¦). Suppose that a rightangled triangle has the property that the lengths of its sides form an arithmetic sequence. Prove that the sides of the triangle are in the ratio 3:4:5.β Example of a bad proof: βIf the sides of the triangle are in the ratio 3:4:5, we could have any multiple of this. e.g. 6, 8, 10. These are still in an arithmetic sequence and satisfy Pythagoras Theorem.β Why is it bad? Weβre supposed to be showing that IF the sides form an arithmetic sequence THEN the sides have ratio 3:4:5. Theyβve just shown the reverse! ? What would a good proof look like? Use our general structure for an algebraic proof! β’ βAssume that the sides form an arithmetic sequence. β’ Model this assumption, ? using algebraic expressions for side lengths that represent all possible arithmetic sequences, e.g. a-b, a, a+b. β’ Do some manipulation using Pythagoras... β’ Therefore, the sides have ratio 3:4:5.β QED QED is short for the latin βQuad erat demonstrandumβ. ? It translates roughly to βWhich had to be ? demonstratedβ. Itβs symbol in maths is β‘ ? I think of a two-digit number. I then reverse the digits. Prove that the difference between the two numbers is a multiple of 9. βLet the number have the digits βabβ. Then the number is 10a + b. If we reverse the digits to get βbaβ, which has the value 10b + a. Subtracting to find the difference: 10a + b β (10b + a) = 9a β 9b = 9(a-b). β‘ 9(a-b) is divisible by 9.β βWithout loss of generalityβ I think of a two-digit number. I then reverse the digits. Prove that the difference between the two numbers is a multiple of 9. βLet the number have the digits βabβ. Then the number is 10a + b. If we reverse the digits to get βbaβ, which has the value 10b + a. Subtracting to find the difference: 10a + b β (10b + a) = 9a β 9b = 9(a-b). β‘ 9(a-b) is divisible by 9.β When we find the difference between two numbers we tend to do the bigger number subtracting the smaller number. Here in our proof we did βππβββππβ. If this difference is positive what have we assumed? That π β₯ π ? If we stated this assumption that π β₯ π, have we made our proof βless generalβ, in the sense that weβve only now proved the property for restricted two-digit numbers where the first digit is greater than (or equal to) the second? No, it doesnβt matter, because if we had say ππ, then after weβd reversed the digits to get 75, we could have subtracted the other way round. The proof still implicitly covers ? all two-digit numbers because the original number and the reversed number are interchangeable when we come to subtract. βWithout loss of generalityβ I think of a two-digit number. I then reverse the digits. Prove that the difference between the two numbers is a multiple of 9. βLet the number have the digits βππβ. With loss of generality, let π > π. Then the number is 10π + π. If we reverse the digits to get βbaβ, which has the value 10b + a. Subtracting to find the difference: 10a + b β (10b + a) = 9a β 9b = 9(a-b). β‘ 9(a-b) is divisible by 9.β ! βWithout loss of generalityβ (or w.l.o.g. for short) means that we have assumed a specific case, but this doesnβt make the proof any less βgeneralβ. Other Types of Proof [Source: JMO] βX and Y play a game in which X starts by choosing a number, which must be either 1 or 2. Y then adds either 1 or 2 and states the total of the two numbers chosen so far. X does likewise, adding either 1 or 2 and stating the total, and so on. The winner is first player to make the total reach (or exceed) 20. i) Explain how X can always win. β¦ For (i), what do we actually have to show? That for any way in which ? Y plays, we can always find a resulting move that guarantees us to win. Games Other Types of Proof Even/Odd You will do this kind of proof next year... βShow that for any integer n, π2 + π is always even.β All integers are either odd or even. 1. If n is odd: n2 is odd because odd x odd = odd. Then odd + odd = even. 2. If n is even: n2 is even because even x even = even. Then even + even = even. ? What weβve done here is split up all numbers according to some property (i.e. odd or even). This made proving our assertion for all integers easier. Summary So what makes a good (and valid!) proof? 1. Use of algebra: a) When appropriate, use variables to represent something that could take any value (e.g. the two digits!) This keeps your argument as general as possible. b) Then use algebra to manipulate these variables based on the question. 2. Keep things general! Avoid making assertions that would ignore certain possibilities. 3. Donβt make leaps of logic Try not to βassumeβ too much. You need to clearly show your steps without making assumptions, which may turn out to be wrong! 4. Avoid circular arguments If the statement to prove is something like βIf X, then show that Yβ, then your proof needs to be something like: βAssume that X, then [manipulation involving X]... And thus Y is true.β You canβt assume Y is true at any point, because thatβs what youβre trying to prove! 5. Consider edge cases Consider unusual cases. e.g. What if the first digit is 0? What if the digits are the same? Does our proof still work? 6. Avoid listing out lots of cases If youβre having to list out lots and lots of cases, then youβre possibly missing ways to narrow down your search. Could you use algebra?