Year 12 proof worksheet answers to download as PDF

These Year 12 proof worksheet answers give a full worked solution to all 21 questions of the proof worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. Let the odd numbers be \(2m+1\) and \(2n+1\). Their sum is \(2m+2n+2=2(m+n+1)\). This is 2 times an integer, so it is even.
2. Let the even number be \(2m\) and the other integer be \(n\). The product is \(2mn=2(mn)\), which is 2 times an integer, so it is even.
3. \((2n+1)^2=4n^2+4n+1=2(2n^2+2n)+1\). This has the form \(2k+1\) with \(k=2n^2+2n\) an integer, so the square is odd.
Answer to Question 2
1. \(n+(n+1)+(n+2)=3n+3=3(n+1)\), a multiple of 3.
2. \(n+(n+1)+(n+2)+(n+3)=4n+6=4(n+1)+2\). This leaves remainder 2 when divided by 4, so it is never a multiple of 4.
3. The product is \(n(n+1)\). Either \(n\) is even or \(n+1\) is even (one of two consecutive integers is even). An even number times an integer is even, so the product is even. Alternatively, check both cases: if \(n=2k\) the product is \(2k(2k+1)\), and if \(n=2k+1\) it is \((2k+1)(2k+2)=2(2k+1)(k+1)\). Both are even.
Answer to Question 3
1. Try \(n=1\): 13, prime. \(n=2\): 17, prime. \(n=3\): 23. \(n=4\): 31. \(n=5\): 41. \(n=6\): 53. \(n=7\): 67. \(n=8\): 83. \(n=9\): 101. \(n=10\): 121 \(=11^2\), which is not prime. So \(n=10\) is a counter-example.
2. Take \(x=-4\). Then \(x^2=16>9\) but \(-4\) is not greater than 3. Counter-example: \(x=-4\).
3. \(2+3=5\) is odd (and 5 is not even). Counter-example: 2 and 3.
4. Take \(a=3\) and \(b=4\): \(\sqrt{9+16}=5\) but \(a+b=7\). Counter-example: \(a=3\), \(b=4\).
Answer to Question 4
1. The last digit of \(n^2\) depends only on the last digit of \(n\). Last digits 0 to 9 give squares ending in \(0,1,4,9,6,5,6,9,4,1\). Every one of these is in the list {0, 1, 4, 5, 6, 9}, so the statement is true.
2. Every integer can be written \(3k\), \(3k+1\) or \(3k+2\). \((3k)^3=27k^3=9(3k^3)\). \((3k+1)^3=27k^3+27k^2+9k+1=9(3k^3+3k^2+k)+1\). \((3k+2)^3=27k^3+54k^2+36k+8=9(3k^3+6k^2+4k)+9-1=9(3k^3+6k^2+4k+1)-1\). These three cases cover every integer, so the statement is true.
Answer to Question 5
1. To prove an identity, start from one side and simplify until you reach the other side. Expand both squares using \((a\pm b)^2=a^2\pm2ab+b^2\). LHS \(=(n^2+6n+9)-(n^2-6n+9)=12n=\) RHS. The \(n^2\) and constant terms cancel, which proves the identity for every value of \(n\).
2. Use the same method, expanding \((2n\pm1)^2\). LHS \(=(4n^2+4n+1)-(4n^2-4n+1)=8n=\) RHS. So the difference of the squares of two consecutive odd numbers is always a multiple of 8.
3. Here it is easier to start from the right-hand side and expand. RHS \(=n(n^2-1)=n^3-n=\) LHS. (Factorising the LHS: \(n^3-n=n(n^2-1)=n(n-1)(n+1)\) by the difference of two squares.) Since \(n-1\), \(n\) and \(n+1\) are three consecutive integers, this also shows that \(n^3-n\) is always divisible by 6.
Answer to Question 6
1. Squaring gives \(2=\dfrac{a^2}{b^2}\), so \(a^2=2b^2\).
2. \(a^2=2b^2\) is even, and the square of an odd number is odd, so \(a\) cannot be odd. Hence \(a\) is even.
3. \(a=2k\) gives \(4k^2=2b^2\), so \(b^2=2k^2\), which is even, so \(b\) is even by the same argument. Now \(a\) and \(b\) are both even, so the fraction \(\dfrac ab\) was not in its simplest form. This contradicts the assumption, so \(\sqrt2\) is irrational.
Answer to Question 7
1. The block is \(n\), \(n+1\), \(n+7\), \(n+8\).
2. Diagonals: \(n+(n+8)=2n+8\) and \((n+1)+(n+7)=2n+8\). They are equal.
3. Total \(=n+(n+1)+(n+7)+(n+8)=4n+16=4(n+4)\). So the total is always a multiple of 4. For example \(9+10+16+17=52=4\times13\).
4. \(4n+16=70\) gives \(4n=54\), \(n=13.5\), not an integer. So no: 70 is not a multiple of 4, so it cannot be the total.
Answer to Question 8
1. The path is the region between the larger square and the lawn. Large square has side \(x+2\), so area \((x+2)^2=x^2+4x+4\). Subtract the lawn \(x^2\): path area \(=4x+4\). This is the required expression.
2. Factorise the expression. \(4x+4=4(x+1)\), which is 4 times an integer. So the area of the path is always a multiple of 4.
3. Form an equation and solve it. \(4x+4=84\) gives \(x=20\). The lawn is 20 m by 20 m; check: \(22^2-20^2=84\) ✓.
4. \(4(x+1)\) is always even (a multiple of 4), so it can never be odd. So the path area is never odd.
Answer to Question 9
1. Each of \(n\) people shakes hands with \(n-1\) others, giving \(n(n-1)\), but every handshake has been counted twice (once by each person). So the total is \(\dfrac{n(n-1)}2\).
2. If \(n\) is even then \(n(n-1)\) is even. If \(n\) is odd then \(n-1\) is even and the product is even. Either way the product is even, so dividing by 2 gives an integer.
3. \(n(n-1)=132\), so \(n^2-n-132=0\), \((n-12)(n+11)=0\). Since \(n>0\), \(n=\) 12.
4. \(n(n-1)=100\). \(10\times9=90\) and \(11\times10=110\), so no integer works. No.
Answer to Question 10
1. Reverse the digits of each number and add. \(37+73=110=11\times10\). \(82+28=110=11\times10\). Both work. Both examples support the claim, but examples alone do not prove it.
2. The reversed number has tens digit \(b\) and units digit \(a\), so it equals \(10b+a\). \((10a+b)+(10b+a)=11a+11b=11(a+b)\), a multiple of 11. Since \(a+b\) is an integer, the sum is always a multiple of 11.
3. Subtract instead of adding. \((10a+b)-(10b+a)=9a-9b=9(a-b)\), a multiple of 9. Since \(a-b\) is an integer, the difference is always a multiple of 9.
4. To show that a statement is false, one counter-example is enough. For 52 and 25: \(52-25=27\), and 27 is not a multiple of 11. Counter-example: 52. So the difference is not always a multiple of 11.
Answer to Question 11
1. \((n+1)^2-n^2=2n+1\).
2. \(2n+1=2(n)+1\) has the form \(2k+1\), so it is always odd.
3. \(0.85(2n+1)\le30\) gives \(2n+1\le35.29\), so \(2n+1\le35\), \(n\le17\). So the largest starting square is \(17\times17\) extended to \(18\times18\).
4. Going from \(k\times k\) to \((k+1)\times(k+1)\) adds the odd number \(2k+1\). Starting with 1 tile, adding 3, then 5, and so on gives \(1+3+5+\dots+(2n-1)=n^2\). Algebraically, the sum of an arithmetic series with \(n\) terms, first term 1 and last term \(2n-1\), is \(\tfrac n2(1+2n-1)=n^2\).
Answer to Question 12
1. Write the fare as \(3+2n\) and look for the form \(2k+1\). Fare \(=3+2n=2n+2+1=2(n+1)+1\), odd. So every taxi fare is an odd number of pounds.
2. Write the fare as \(5+4n\). Fare \(=5+4n=4n+4+1=2(2n+2)+1\), odd. So every bus fare is also an odd number of pounds.
3. Subtract the two fares and factorise. Odd minus odd is even: \((5+4n)-(3+2n)=2n+2=2(n+1)\). Since \(n+1\) is an integer, the difference is always even.
4. Set the fares equal, which is the same as setting the difference equal to 0. \(2n+2=0\) gives \(n=-1\), which is impossible, so the fares are never equal (the bus is always dearer). The difference is always at least £2.
Answer to Question 13
1. Let the number be \(n\) and write each step algebraically. \(\dfrac{2n+6}{2}-n=n+3-n=3\). The original number cancels, so the answer is always 3, whatever number is chosen.
2. Write each step algebraically; since \(n\ne0\) we can divide by \(n\). \(\dfrac{n^2+n}{n}-n=n+1-n=1\). The answer is always 1.
3. Writing a three-digit number twice is the same as multiplying it by 1000 and adding the number itself. A three-digit number \(N\) repeated is \(1000N+N=1001N\). \(1001=7\times11\times13\), so \(1001N\) is divisible by 7, 11 and 13. For example \(247247=1001\times247\).
Answer to Question 14
1. The total is \(1+2+\dots+n=\dfrac n2(1+n)\), the sum of an arithmetic series.
2. \(n(n+1)\) is a product of consecutive integers so it is even, and dividing by 2 gives an integer.
3. \(\tfrac12n(n+1)>500\) gives \(n(n+1)>1000\). \(31\times32=992<1000\) and \(32\times33=1056>1000\). So on day 32 (total £528).
4. \(\tfrac12n(n+1)+\tfrac12(n+1)(n+2)=\tfrac12(n+1)(n+n+2)=(n+1)^2\).
Answer to Question 15
1. \(n=4\): \(9\), \(40\), \(41\). \(9^2+40^2=81+1600=1681=41^2\). It works.
2. \((2n+1)^2+(2n^2+2n)^2=4n^2+4n+1+4n^4+8n^3+4n^2=4n^4+8n^3+8n^2+4n+1\). And \((2n^2+2n+1)^2=4n^4+8n^3+4n^2+4n^2+4n+1=4n^4+8n^3+8n^2+4n+1\). They are equal, so the rule always works.
3. \(2n^2+2n+1=85\) gives \(n^2+n-42=0\), \((n+7)(n-6)=0\), \(n=6\). The triple is \(13\), \(84\), \(85\): \(169+7056=7225=85^2\).
4. The 3-4-5 triangle has perimeter 12. Multiply every side by 5 to get 15, 20, 25 with perimeter 60.
Answer to Question 16
1. The product \(p_1p_2\cdots p_k\) is divisible by every \(p_i\), so adding 1 gives remainder 1 on division by each \(p_i\).
2. \(N>1\), so it is either prime itself (and \(N\) is bigger than every prime on the list, so it is a new prime) or it has a prime factor. That prime factor cannot be any \(p_i\) because dividing by \(p_i\) leaves remainder 1. So there is a prime not on the list.
3. The list was assumed to contain all primes, but we have found a prime that is not on it. This is a contradiction, so the assumption is false and there are infinitely many primes.
4. \(N=2\times3\times5\times7\times11\times13+1=30031=59\times509\). So \(N\) is not prime. This does not break the argument: the prime factors 59 and 509 are not on the list, which is exactly what the proof predicts.
Answer to Question 17
1. Factorise the expression. True. \(n^2-n=n(n-1)\), a product of consecutive integers, which is always even. So the statement is true for every integer \(n\).
2. Use the result of part (a) with \(n^2+n\). True. \(n^2+n=n(n+1)\) is even, so adding 1 gives an odd number. So the statement is true for every integer \(n\).
3. Try small values of \(n\): \(n=1\) gives 2, which is not a multiple of 3 either. False. \(n=2\) gives \(8+2=10\), not a multiple of 3. So the statement is false.
4. Try small values: \(n=1\) gives 3 and \(n=2\) gives 5, both prime. False. \(n=3\) gives \(2^3+1=9=3\times3\), which is not prime. So the statement is false.
Answer to Question 18
1. Write the three integers as \(n-1\), \(n\) and \(n+1\). \(\dfrac{(n-1)+n+(n+1)}{3}=\dfrac{3n}{3}=n\). So the mean is the middle integer.
2. Expand each square and collect terms. \((n-1)^2+n^2+(n+1)^2=(n^2-2n+1)+n^2+(n^2+2n+1)=3n^2+2\), which is 2 more than the multiple of 3 given by \(3n^2\). So the sum always leaves remainder 2 when divided by 3.
3. Write the integers as \(n\) and \(n+1\). \(n^2+(n+1)^2=2n^2+2n+1=2(n^2+n)+1\), odd. So the sum is always odd.
4. Use the expression from part (c) and solve the quadratic. \(2n^2+2n+1=421\) gives \(n^2+n-210=0\), \((n+15)(n-14)=0\), so \(n=14\). The integers are 14 and 15: \(196+225=421\). Using the negative root \(n=-15\) gives the pair \(-15\) and \(-14\), which also works.
Answer to Question 19
1. \(x^2-6x+10=(x-3)^2+1\). A square is at least 0, so the expression is at least 1, which is positive.
2. \(a^2+b^2-2ab=(a-b)^2\ge0\), so \(a^2+b^2\ge2ab\).
3. Since \(x>0\), multiply by \(x\) without reversing the inequality: \(x^2+1\ge2x\), i.e. \(x^2-2x+1\ge0\), i.e. \((x-1)^2\ge0\), which is true. Reversing the steps proves the original.
4. The inequality in part (c) is exactly this statement. Equality holds when \((x-1)^2=0\), i.e. \(x=1\).
Answer to Question 20
1. Every integer is even (\(2m\)) or odd (\(2m+1\)).
2. \(n^2\) leaves remainder 0 or 1 on division by 4. The remainders 2 and 3 never occur, so a square cannot have the form \(4k+2\) or \(4k+3\).
3. \(1\,000\,003=4\times250\,000+3\), which has the form \(4k+3\), so it is not a square.
4. \(x^2=4y+3\) would mean \(x^2\) has remainder 3 on division by 4, which is impossible. So there are no integer solutions.
Answer to Question 21
1. Assume \(x+y=r\) is rational. Then \(y=r-x\), a difference of two rational numbers, which is rational. This contradicts \(y\) being irrational. So \(x+y\) is irrational.
2. Assume \(\sqrt2+\sqrt3=r\) is rational. Squaring gives \(5+2\sqrt6=r^2\), so \(\sqrt6=\dfrac{r^2-5}2\), which is rational. This contradicts the given fact that \(\sqrt6\) is irrational. So \(\sqrt2+\sqrt3\) is irrational.
3. Assume \(\log_23=\dfrac pq\) with positive integers \(p,q\). Then \(2^{p/q}=3\), so \(2^p=3^q\). The left side is even and the right side is odd (for \(q\ge1\)), a contradiction. So \(\log_23\) is irrational.

How to use these Year 12 proof worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the proof lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 12 proof worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 12 proof worksheet answers show every step, so nothing is left unexplained.