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bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-1-1-proof-by-contradiction/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":41,"initialQuestionWindow":[{"id":"4470672a-f3bb-48d3-96e8-994a0a98913b","revisionId":"ddc5b72f-3a58-4182-b14b-c84b5530b9b9","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"hard","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"3ca1a422-2936-48fa-a461-92e936bd292b","label":"a","content":"Show that $\\sqrt{11}$ is irrational by using a proof by contradiction.","markscheme":"- Assume $\\sqrt{11}$ is rational, such that $\\sqrt{11} = \\frac{m}{n}$ where $m, n$ are integers and $\\frac{m}{n}$ is in its lowest terms. B1\n- Squaring gives $11 = \\frac{m^2}{n^2} \\Rightarrow 11n^2 = m^2$ M1\n- $m^2$ is a multiple of 11, which implies $m$ is a multiple of 11. A1\n- Let $m = 11k$. Then $11n^2 = (11k)^2 = 121k^2 \\Rightarrow n^2 = 11k^2$ M1\n- $n^2$ is a multiple of 11, which implies $n$ is a multiple of 11. A1\n- As $m$ and $n$ both have a common factor of 11, the initial assumption that the fraction is in lowest terms is contradicted. Hence, $\\sqrt{11}$ is irrational. A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Assume $\\sqrt{11} = m/n$ where $m, n$ have no common factors.","code":"B"},{"id":"p2","marks":1,"description":"Square both sides and rearrange to get $m^2 = 11n^2$.","code":"M"},{"id":"p3","marks":1,"description":"Correctly deduce that $m$ is a multiple of 11.","code":"A"},{"id":"p4","marks":1,"description":"Substitute $m = 11k$ and simplify to get $n^2 = 11k^2$.","code":"M"},{"id":"p5","marks":1,"description":"Correctly deduce that $n$ is a multiple of 11.","code":"A"},{"id":"p6","marks":1,"description":"Conclude that having a common factor of 11 is a contradiction.","code":"A"}],"finalAnswers":["Contradiction shown: common factor of 11 exists"],"generatedAt":"2026-05-06T14:03:29.312Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"coprime_contradiction","label":"Coprime Integers Approach","steps":[{"type":"text","order":0,"title":"Define the contradiction assumption","content":"To prove that $\\sqrt{11}$ is irrational by contradiction, we begin by assuming the opposite. \n\nAssume that $\\sqrt{11}$ is a **rational** number. By definition, a rational number can be written as a fraction:\n\n$$\\sqrt{11} = \\frac{m}{n}$$\n\nwhere $m$ and $n$ are non-zero integers that are **coprime** (they have no common factors other than 1, meaning the fraction $\\frac{m}{n}$ is in its simplest form).\n\n*Award **1 mark** (B1) for setting up this initial assumption.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sqrt{11} = \\frac{\\htmlData{anchor=eq1-m}{m}}{\\htmlData{anchor=eq1-n}{n}}$"},{"latex":"$$11 = \\frac{\\htmlData{anchor=eq2-m2}{m^2}}{\\htmlData{anchor=eq2-n2}{n^2}}$"},{"latex":"$$\\htmlData{anchor=eq3-m2}{m^2} = 11\\htmlData{anchor=eq3-n2}{n^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq1-m"},{"color":"purple","style":"text-color","anchor":"eq1-n"}],"connections":[{"to":"eq2-m2","from":"eq1-m","color":"sky","label":"square","style":"solid"},{"to":"eq2-n2","from":"eq1-n","color":"purple","label":"square","style":"solid"},{"to":"eq3-m2","from":"eq2-m2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Square and rearrange","description":"Now, square both sides to remove the square root and rearrange the equation to isolate $m^2$.\n\n*Award **1 mark** (M1) for squaring and rearranging to obtain $m^2 = 11n^2$.*"},{"type":"text","order":2,"title":"Analyze divisibility of m","content":"From the equation $m^2 = 11n^2$, we can see that $m^2$ must be a multiple of 11.\n\nSince 11 is a prime number, if 11 divides $m^2$, then 11 must also divide $m$.\n\nTherefore, we can express $m$ as:\n\n$$m = 11k$$\n\nwhere $k$ is an integer.\n\n*Award **1 mark** (A1) for deducing that $m$ is a multiple of 11.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=sub-m}{m} = 11k$"},{"latex":"$$\\htmlData{anchor=sub-lhs}{m^2} = 11n^2$"},{"latex":"$$(\\htmlData{anchor=sub-rhs}{11k})^2 = 11n^2$"},{"latex":"$$121k^2 = 11n^2$"},{"latex":"$$n^2 = 11k^2$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sub-m"},{"color":"amber","style":"text-color","anchor":"sub-rhs"}],"connections":[{"to":"sub-rhs","from":"sub-m","color":"amber","label":"substitute","style":"solid"}]},"order":3,"title":"Substitute and analyze n","description":"Substitute $m = 11k$ back into our equation $m^2 = 11n^2$ and simplify.\n\nDividing both sides of $121k^2 = 11n^2$ by 11 gives:\n$$n^2 = 11k^2$$\n\n*Award **1 mark** (M1) for substituting $m = 11k$ and simplifying to $n^2 = 11k^2$.*"},{"type":"text","order":4,"title":"Analyze divisibility of n","content":"From the equation $n^2 = 11k^2$, we can see that $n^2$ is a multiple of 11.\n\nSince 11 is a prime number, if 11 divides $n^2$, then 11 must also divide $n$.\n\nTherefore, $n$ is also a multiple of 11.\n\n*Award **1 mark** (A1) for deducing that $n$ is a multiple of 11.*"},{"type":"text","order":5,"title":"Conclude the contradiction","content":"We have shown that both $m$ and $n$ are multiples of 11, meaning they share a common factor of 11.\n\nThis directly contradicts our initial assumption that $\\frac{m}{n}$ is in its simplest form (where $m$ and $n$ are coprime integers sharing no common factors other than 1).\n\nSince the assumption leads to a contradiction, it must be false. Therefore, $\\sqrt{11}$ cannot be rational, which proves that **$\\sqrt{11}$ is irrational**.\n\n*Award **1 mark** (A1) for completing the contradiction argument and writing a concluding statement.*"}]},{"id":"prime_factorization","label":"Unique Prime Factorization Approach","steps":[{"type":"text","order":0,"title":"State the initial assumption","content":"To prove that $\\sqrt{11}$ is irrational using contradiction, we begin by assuming the opposite: that $\\sqrt{11}$ is a **rational** number. \n\nBy definition, a rational number can be expressed as a fraction of two integers:\n\n$$\\sqrt{11} = \\frac{m}{n}$$\n\nwhere $m$ and $n$ are non-zero integers.\n\n*Note: In this unique prime factorization approach, we do not even need to assume that the fraction is in its lowest terms!*"},{"type":"text","order":1,"title":"Square and rearrange","content":"Next, let's eliminate the square root by squaring both sides of the equation:\n\n$$11 = \\frac{m^2}{n^2}$$\n\nMultiplying both sides by $n^2$ gives:\n\n$$m^2 = 11n^2$$\n\nAccording to the **Fundamental Theorem of Arithmetic**, every integer greater than $1$ has a unique prime factorization. We will now count how many times the prime number $11$ appears in the prime factorization of both sides of this equation."},{"type":"text","order":2,"title":"Analyze the left-hand side","content":"Let's find the multiplicity (the exponent) of the prime factor $11$ on the left-hand side, $m^2$.\n\nSuppose the prime factorization of $m$ contains the prime factor $11$ exactly $k$ times (where $k \\ge 0$). When we square $m$ to get $m^2$, every prime exponent in the factorization is doubled.\n\nTherefore, the number of times $11$ appears in the prime factorization of $m^2$ is:\n\n$$\\text{Multiplicity of 11 in } m^2 = 2k$$\n\nSince $2k$ is a multiple of $2$, the prime factor $11$ must appear an **even number of times** on the left-hand side."},{"type":"text","order":3,"title":"Analyze the right-hand side","content":"Now let's find the multiplicity of the prime factor $11$ on the right-hand side, $11n^2$.\n\nSuppose the prime factorization of $n$ contains the prime factor $11$ exactly $j$ times (where $j \\ge 0$). When we square $n$ to get $n^2$, the prime factor $11$ appears $2j$ times (an even number of times).\n\nWhen we multiply $n^2$ by $11$, we add one more factor of $11$ to the factorization:\n\n$$\\text{Multiplicity of 11 in } 11n^2 = 2j + 1$$\n\nSince $2j + 1$ is always odd, the prime factor $11$ must appear an **odd number of times** on the right-hand side."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$11 \\text{ appears } \\htmlData{anchor=lhs-mult}{\\color{@sky}{2k}} \\text{ times in } \\htmlData{anchor=lhs}{\\color{@sky}{m^2}}$"},{"latex":"$$11 \\text{ appears } \\htmlData{anchor=rhs-mult}{\\color{@amber}{2j+1}} \\text{ times in } \\htmlData{anchor=rhs}{\\color{@amber}{11n^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"lhs-mult"},{"color":"amber","style":"text-color","anchor":"rhs-mult"}],"connections":[{"to":"lhs-mult","from":"lhs","color":"sky","label":"even power","style":"solid"},{"to":"rhs-mult","from":"rhs","color":"amber","label":"odd power","style":"solid"}]},"order":4,"title":"Compare the multiplicities","description":"Because the two sides are equal ($m^2 = 11n^2$), they represent the same integer and must have the exact same unique prime factorization."},{"type":"text","order":5,"title":"Reach a contradiction","content":"We have established a direct contradiction:\n- The left-hand side ($m^2$) contains an **even** number of prime factors of $11$.\n- The right-hand side ($11n^2$) contains an **odd** number of prime factors of $11$.\n\nBy the **Fundamental Theorem of Arithmetic**, an integer cannot have two different prime factorizations. Therefore, the equation $m^2 = 11n^2$ has no integer solutions for non-zero $m$ and $n$.\n\nThis means our initial assumption that $\\sqrt{11}$ is rational is false. Thus, **$\\sqrt{11}$ is irrational**."}]}],"generatedAt":"2026-06-09T04:40:08.509Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"3ca1a422-2936-48fa-a461-92e936bd292b:coprime_contradiction","label":"Coprime Integers Approach","steps":[{"type":"text","order":0,"title":"Define the contradiction assumption","content":"To prove that $\\sqrt{11}$ is irrational by contradiction, we begin by assuming the opposite. \n\nAssume that $\\sqrt{11}$ is a **rational** number. By definition, a rational number can be written as a fraction:\n\n$$\\sqrt{11} = \\frac{m}{n}$$\n\nwhere $m$ and $n$ are non-zero integers that are **coprime** (they have no common factors other than 1, meaning the fraction $\\frac{m}{n}$ is in its simplest form).\n\n*Award **1 mark** (B1) for setting up this initial assumption.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sqrt{11} = \\frac{\\htmlData{anchor=eq1-m}{m}}{\\htmlData{anchor=eq1-n}{n}}$"},{"latex":"$$11 = \\frac{\\htmlData{anchor=eq2-m2}{m^2}}{\\htmlData{anchor=eq2-n2}{n^2}}$"},{"latex":"$$\\htmlData{anchor=eq3-m2}{m^2} = 11\\htmlData{anchor=eq3-n2}{n^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq1-m"},{"color":"purple","style":"text-color","anchor":"eq1-n"}],"connections":[{"to":"eq2-m2","from":"eq1-m","color":"sky","label":"square","style":"solid"},{"to":"eq2-n2","from":"eq1-n","color":"purple","label":"square","style":"solid"},{"to":"eq3-m2","from":"eq2-m2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Square and rearrange","description":"Now, square both sides to remove the square root and rearrange the equation to isolate $m^2$.\n\n*Award **1 mark** (M1) for squaring and rearranging to obtain $m^2 = 11n^2$.*"},{"type":"text","order":2,"title":"Analyze divisibility of m","content":"From the equation $m^2 = 11n^2$, we can see that $m^2$ must be a multiple of 11.\n\nSince 11 is a prime number, if 11 divides $m^2$, then 11 must also divide $m$.\n\nTherefore, we can express $m$ as:\n\n$$m = 11k$$\n\nwhere $k$ is an integer.\n\n*Award **1 mark** (A1) for deducing that $m$ is a multiple of 11.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=sub-m}{m} = 11k$"},{"latex":"$$\\htmlData{anchor=sub-lhs}{m^2} = 11n^2$"},{"latex":"$$(\\htmlData{anchor=sub-rhs}{11k})^2 = 11n^2$"},{"latex":"$$121k^2 = 11n^2$"},{"latex":"$$n^2 = 11k^2$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sub-m"},{"color":"amber","style":"text-color","anchor":"sub-rhs"}],"connections":[{"to":"sub-rhs","from":"sub-m","color":"amber","label":"substitute","style":"solid"}]},"order":3,"title":"Substitute and analyze n","description":"Substitute $m = 11k$ back into our equation $m^2 = 11n^2$ and simplify.\n\nDividing both sides of $121k^2 = 11n^2$ by 11 gives:\n$$n^2 = 11k^2$$\n\n*Award **1 mark** (M1) for substituting $m = 11k$ and simplifying to $n^2 = 11k^2$.*"},{"type":"text","order":4,"title":"Analyze divisibility of n","content":"From the equation $n^2 = 11k^2$, we can see that $n^2$ is a multiple of 11.\n\nSince 11 is a prime number, if 11 divides $n^2$, then 11 must also divide $n$.\n\nTherefore, $n$ is also a multiple of 11.\n\n*Award **1 mark** (A1) for deducing that $n$ is a multiple of 11.*"},{"type":"text","order":5,"title":"Conclude the contradiction","content":"We have shown that both $m$ and $n$ are multiples of 11, meaning they share a common factor of 11.\n\nThis directly contradicts our initial assumption that $\\frac{m}{n}$ is in its simplest form (where $m$ and $n$ are coprime integers sharing no common factors other than 1).\n\nSince the assumption leads to a contradiction, it must be false. Therefore, $\\sqrt{11}$ cannot be rational, which proves that **$\\sqrt{11}$ is irrational**.\n\n*Award **1 mark** (A1) for completing the contradiction argument and writing a concluding statement.*"}]},{"id":"3ca1a422-2936-48fa-a461-92e936bd292b:prime_factorization","label":"Unique Prime Factorization Approach","steps":[{"type":"text","order":0,"title":"State the initial assumption","content":"To prove that $\\sqrt{11}$ is irrational using contradiction, we begin by assuming the opposite: that $\\sqrt{11}$ is a **rational** number. \n\nBy definition, a rational number can be expressed as a fraction of two integers:\n\n$$\\sqrt{11} = \\frac{m}{n}$$\n\nwhere $m$ and $n$ are non-zero integers.\n\n*Note: In this unique prime factorization approach, we do not even need to assume that the fraction is in its lowest terms!*"},{"type":"text","order":1,"title":"Square and rearrange","content":"Next, let's eliminate the square root by squaring both sides of the equation:\n\n$$11 = \\frac{m^2}{n^2}$$\n\nMultiplying both sides by $n^2$ gives:\n\n$$m^2 = 11n^2$$\n\nAccording to the **Fundamental Theorem of Arithmetic**, every integer greater than $1$ has a unique prime factorization. We will now count how many times the prime number $11$ appears in the prime factorization of both sides of this equation."},{"type":"text","order":2,"title":"Analyze the left-hand side","content":"Let's find the multiplicity (the exponent) of the prime factor $11$ on the left-hand side, $m^2$.\n\nSuppose the prime factorization of $m$ contains the prime factor $11$ exactly $k$ times (where $k \\ge 0$). When we square $m$ to get $m^2$, every prime exponent in the factorization is doubled.\n\nTherefore, the number of times $11$ appears in the prime factorization of $m^2$ is:\n\n$$\\text{Multiplicity of 11 in } m^2 = 2k$$\n\nSince $2k$ is a multiple of $2$, the prime factor $11$ must appear an **even number of times** on the left-hand side."},{"type":"text","order":3,"title":"Analyze the right-hand side","content":"Now let's find the multiplicity of the prime factor $11$ on the right-hand side, $11n^2$.\n\nSuppose the prime factorization of $n$ contains the prime factor $11$ exactly $j$ times (where $j \\ge 0$). When we square $n$ to get $n^2$, the prime factor $11$ appears $2j$ times (an even number of times).\n\nWhen we multiply $n^2$ by $11$, we add one more factor of $11$ to the factorization:\n\n$$\\text{Multiplicity of 11 in } 11n^2 = 2j + 1$$\n\nSince $2j + 1$ is always odd, the prime factor $11$ must appear an **odd number of times** on the right-hand side."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$11 \\text{ appears } \\htmlData{anchor=lhs-mult}{\\color{@sky}{2k}} \\text{ times in } \\htmlData{anchor=lhs}{\\color{@sky}{m^2}}$"},{"latex":"$$11 \\text{ appears } \\htmlData{anchor=rhs-mult}{\\color{@amber}{2j+1}} \\text{ times in } \\htmlData{anchor=rhs}{\\color{@amber}{11n^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"lhs-mult"},{"color":"amber","style":"text-color","anchor":"rhs-mult"}],"connections":[{"to":"lhs-mult","from":"lhs","color":"sky","label":"even power","style":"solid"},{"to":"rhs-mult","from":"rhs","color":"amber","label":"odd power","style":"solid"}]},"order":4,"title":"Compare the multiplicities","description":"Because the two sides are equal ($m^2 = 11n^2$), they represent the same integer and must have the exact same unique prime factorization."},{"type":"text","order":5,"title":"Reach a contradiction","content":"We have established a direct contradiction:\n- The left-hand side ($m^2$) contains an **even** number of prime factors of $11$.\n- The right-hand side ($11n^2$) contains an **odd** number of prime factors of $11$.\n\nBy the **Fundamental Theorem of Arithmetic**, an integer cannot have two different prime factorizations. Therefore, the equation $m^2 = 11n^2$ has no integer solutions for non-zero $m$ and $n$.\n\nThis means our initial assumption that $\\sqrt{11}$ is rational is false. Thus, **$\\sqrt{11}$ is irrational**."}]}],"generatedAt":"2026-06-09T04:40:21.901Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"d071f313-7932-4ad0-aab3-9f797eda3853","revisionId":"f5c63d68-a48b-45ca-9d8b-e8b1af9c597f","questionNumber":"6","questionType":"LA","specification":"","difficulty":3,"difficultyBand":"hard","totalMarks":6,"sourceTitle":"1.1 Proof by Contradiction","calculatorUse":null,"isCalculation":true,"parts":[{"id":"387555c6-3123-4682-b855-2b9268f84065","label":"0","content":"Prove by contradiction that there are infinitely many prime numbers","markscheme":"- Assume there are a finite number of primes. B1\n- Let $x$ be the product of all prime numbers: $x = p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n$ M1\n- Consider $y = x + 1$ M1\n- Explain that $y$ has no prime factors from $p_1$ to $p_n$ as there would always be a remainder of 1. A1\n- State that either $y$ is prime or $y$ has a prime factor not in the original list. M1\n- Conclude that the assumption of a finite number of primes is incorrect. A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Assumes there is a finite set of prime numbers.","code":"B"},{"id":"p2","marks":1,"description":"Defines a number $x$ as the product of all primes in the finite set.","code":"M"},{"id":"p3","marks":1,"description":"Constructs a new number $y = (p_1 p_2 \\dots p_n) + 1$.","code":"M"},{"id":"p4","marks":1,"description":"Argues that $y$ is not divisible by any of the primes in the finite set (remainder 1).","code":"A"},{"id":"p5","marks":1,"description":"Deduces that $y$ must be prime or have a prime factor not in the set.","code":"M"},{"id":"p6","marks":1,"description":"Concludes that there must be infinitely many primes.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T03:40:14.493Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"euclids_proof","label":"Euclid's Classical Proof","steps":[{"type":"text","order":0,"title":"Assume finite primes","content":"To prove this by contradiction, we begin by assuming the opposite of what we want to show.\n\nAssume that there is a **finite** number of prime numbers. We can list all of them in a complete set:\n\n$$\\{p_1, p_2, p_3, \\dots, p_n\\}$$\n\nwhere $p_1 = 2$, $p_2 = 3$, and $p_n$ represents the largest prime number in existence."},{"type":"text","order":1,"title":"Define the product","content":"Next, let $x$ be the product of all the prime numbers in our finite list:\n\n$$x = p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n$$"},{"type":"text","order":2,"title":"Construct a new integer","content":"Now, we define a new integer, $y$, by adding $1$ to this product:\n\n$$y = x + 1 = (p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n) + 1$$"},{"type":"text","order":3,"title":"Analyze the remainder","content":"Consider what happens when we divide $y$ by any of the prime numbers from our list, $p_i$:\n\n$$\\frac{y}{p_i} = \\frac{p_1 \\times p_2 \\times \\dots \\times p_n}{p_i} + \\frac{1}{p_i}$$\n\nBecause $p_i$ is a factor of the product $x$, the first term divides perfectly to leave an integer, but we are left with a remainder of $1$ from the second term.\n\nTherefore, $y$ is not divisible by any of the primes in our complete list $p_1, p_2, \\dots, p_n$."},{"type":"text","order":4,"title":"Find the contradiction","content":"By the Fundamental Theorem of Arithmetic, every integer greater than $1$ must either be prime itself or be divisible by prime numbers.\n\nThis leaves us with two possibilities for $y$:\n1. $y$ is itself a prime number. Since $y > p_n$, this would be a new prime not in our list.\n2. $y$ is a composite number and must have a prime factor. Since none of the primes in our list divide $y$, this prime factor must be a new prime not in our list.\n\nIn both cases, we have established the existence of a prime number that was not included in our initial 'complete' set."},{"type":"text","order":5,"title":"State the final conclusion","content":"This directly contradicts our initial assumption that the set $\\{p_1, p_2, \\dots, p_n\\}$ contains all prime numbers.\n\nTherefore, our assumption that the number of primes is finite must be false.\n\nConsequently, there are **infinitely many** prime numbers."}]}],"generatedAt":"2026-06-09T09:23:23.225Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"387555c6-3123-4682-b855-2b9268f84065:euclids_proof","label":"Euclid's Classical Proof","steps":[{"type":"text","order":0,"title":"Assume finite primes","content":"To prove this by contradiction, we begin by assuming the opposite of what we want to show.\n\nAssume that there is a **finite** number of prime numbers. We can list all of them in a complete set:\n\n$$\\{p_1, p_2, p_3, \\dots, p_n\\}$$\n\nwhere $p_1 = 2$, $p_2 = 3$, and $p_n$ represents the largest prime number in existence."},{"type":"text","order":1,"title":"Define the product","content":"Next, let $x$ be the product of all the prime numbers in our finite list:\n\n$$x = p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n$$"},{"type":"text","order":2,"title":"Construct a new integer","content":"Now, we define a new integer, $y$, by adding $1$ to this product:\n\n$$y = x + 1 = (p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n) + 1$$"},{"type":"text","order":3,"title":"Analyze the remainder","content":"Consider what happens when we divide $y$ by any of the prime numbers from our list, $p_i$:\n\n$$\\frac{y}{p_i} = \\frac{p_1 \\times p_2 \\times \\dots \\times p_n}{p_i} + \\frac{1}{p_i}$$\n\nBecause $p_i$ is a factor of the product $x$, the first term divides perfectly to leave an integer, but we are left with a remainder of $1$ from the second term.\n\nTherefore, $y$ is not divisible by any of the primes in our complete list $p_1, p_2, \\dots, p_n$."},{"type":"text","order":4,"title":"Find the contradiction","content":"By the Fundamental Theorem of Arithmetic, every integer greater than $1$ must either be prime itself or be divisible by prime numbers.\n\nThis leaves us with two possibilities for $y$:\n1. $y$ is itself a prime number. Since $y > p_n$, this would be a new prime not in our list.\n2. $y$ is a composite number and must have a prime factor. Since none of the primes in our list divide $y$, this prime factor must be a new prime not in our list.\n\nIn both cases, we have established the existence of a prime number that was not included in our initial 'complete' set."},{"type":"text","order":5,"title":"State the final conclusion","content":"This directly contradicts our initial assumption that the set $\\{p_1, p_2, \\dots, p_n\\}$ contains all prime numbers.\n\nTherefore, our assumption that the number of primes is finite must be false.\n\nConsequently, there are **infinitely many** prime numbers."}]}],"generatedAt":"2026-06-09T09:23:32.752Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"242514c0-179f-43d9-8f6f-6909d182f51f","revisionId":"ddf98bc4-2f1f-4699-b6c7-8af48e106b1e","questionNumber":"18","questionType":"LA","specification":"","difficulty":1,"difficultyBand":"hard","totalMarks":6,"sourceTitle":"Question Set 1: Algebraic Methods","calculatorUse":null,"isCalculation":true,"parts":[{"id":"28f0d98f-3363-4b89-8b16-46c25504862b","label":"","content":"Prove by contradiction that there are infinitely many prime numbers","markscheme":"- Assume there is a finite number of prime numbers: $p_1, p_2, p_3, \\dots, p_n$ B1\n- Construct a new number $N = (p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n) + 1$ M1\n- Explain that dividing $N$ by any of the primes $p_1, \\dots, p_n$ leaves a remainder of 1. M1\n- Deduce that $N$ must either be prime itself or have a prime factor not in the original list. A1\n- State that this contradicts the assumption that the list contained all prime numbers. A1\n- Conclude that there must be infinitely many prime numbers. A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Starts with the assumption that there is a finite set of primes.","code":"B"},{"id":"p2","marks":1,"description":"Correctly constructs the number $N$ (product of primes plus 1).","code":"M"},{"id":"p3","marks":1,"description":"Explains the remainder when $N$ is divided by any prime in the set.","code":"M"},{"id":"p4","marks":1,"description":"Deduces that $N$ has a prime factor not in the set.","code":"A"},{"id":"p5","marks":1,"description":"Identifies the contradiction.","code":"A"},{"id":"p6","marks":1,"description":"Provides a complete concluding statement.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T02:36:06.047Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"euclid_product","label":"Euclid's Product Construction","steps":[{"type":"text","order":0,"title":"State the assumption","content":"To prove this by contradiction, we start by assuming the opposite of what we want to prove.\n\nAssume that there is a **finite** number of prime numbers. We can write them all down as a complete list:\n\n$$p_1, p_2, p_3, \\dots, p_n$$\n\nwhere $p_n$ is the largest prime number in existence.\n\n*This step earns the first mark ($B1$) for setting up the correct assumption for contradiction.*"},{"type":"text","order":1,"title":"Construct a new integer","content":"Using Euclid's product construction, we define a new integer $N$ by multiplying all of the prime numbers in our list together and then adding $1$:\n\n$$N = (p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n) + 1$$\n\n*This earns the second mark ($M1$) for constructing the test number $N$.*"},{"type":"text","order":2,"title":"Analyze divisibility","content":"Let's check if any of the prime numbers from our assumed list can divide $N$ exactly.\n\nIf we divide $N$ by any prime $p_i$ from our list, we get:\n\n$$\\frac{N}{p_i} = \\left(\\frac{p_1 \\times p_2 \\times \\dots \\times p_n}{p_i}\\right) + \\frac{1}{p_i}$$\n\nSince $p_i$ is a factor in the numerator of the first term, that division results in an integer. However, the term $\\frac{1}{p_i}$ remains.\n\nTherefore, dividing $N$ by any of the primes $p_1, \\dots, p_n$ leaves a **remainder of $1$**.\n\n*This earns the third mark ($M1$) for establishing the remainder of 1.*"},{"type":"text","order":3,"title":"Deduce prime factor properties","content":"Every integer greater than $1$ must either be a prime number itself or be divisible by a prime number.\n\nSince none of the primes in our finite list $p_1, \\dots, p_n$ can divide $N$ without a remainder, we are left with two possibilities:\n1. $N$ is prime itself (and because $N$ is greater than any $p_i$, it is a new prime not in our list).\n2. $N$ has a prime factor that is not in our original list.\n\n*This earns the fourth mark ($A1$) for deducing that $N$ must either be prime or have a new prime factor.*"},{"type":"text","order":4,"title":"Identify the contradiction","content":"Both possibilities show that there exists a prime number that was not included in our initial list of primes.\n\nThis directly **contradicts** our initial assumption that our list $p_1, p_2, \\dots, p_n$ contained all prime numbers.\n\n*This earns the fifth mark ($A1$) for identifying the contradiction.*"},{"type":"text","order":5,"title":"State the final conclusion","content":"Since our assumption of a finite number of primes leads to a logical contradiction, the assumption must be false.\n\nTherefore, there are **infinitely many prime numbers**.\n\n*This earns the final mark ($A1$) for the complete conclusion.*"}]},{"id":"factorial_construction","label":"Factorial Construction","steps":[{"type":"text","order":0,"title":"Assume finite primes","content":"To prove the statement by contradiction, we start by assuming the opposite of what we want to prove.\n\nAssume that the set of prime numbers is **finite**. Let this complete list of prime numbers be represented as:\n\n$$\\{p_1, p_2, p_3, \\dots, p\\}$$\n\nwhere $p$ is the **largest prime number** in existence.\n\n*This step earns **B1** for setting up the proof by contradiction with a finite list of primes.*"},{"type":"text","order":1,"title":"Construct the test number","content":"Using our assumption, we construct a new integer, $N$, using the factorial of our largest prime $p$:\n\n$$N = p! + 1$$\n\nRecall that the factorial of $p$ is the product of all positive integers from $1$ up to $p$:\n\n$$p! = p \\times (p - 1) \\times (p - 2) \\times \\dots \\times 2 \\times 1$$\n\n*This step earns **M1** for constructing the value $N = p! + 1$.*"},{"type":"text","order":2,"title":"Analyze divisibility","content":"Let's consider what happens when we divide $N$ by any prime number $q$ from our assumed finite list.\n\nSince $p$ is the largest prime in our list, any prime $q$ must satisfy $q \\le p$. Therefore, $q$ is a factor of $p!$:\n\n$$p! = q \\times K$$ \n\nfor some integer $K$. Substitute this back into our expression for $N$:\n\n$$N = qK + 1$$\n\nDividing $N$ by $q$ gives:\n\n$$\\frac{N}{q} = K + \\frac{1}{q}$$\n\nThis shows that dividing $N$ by any of the primes in our finite list always leaves a **remainder of 1**.\n\n*This step earns **M1** for showing that dividing $N$ by any of the assumed primes leaves a remainder of 1.*"},{"type":"text","order":3,"title":"Find the contradiction","content":"By the Fundamental Theorem of Arithmetic, the integer $N > 1$ must either be prime itself or have a prime factor.\n\n- If $N$ is prime, then since $N = p! + 1 > p$, we have found a prime number greater than our supposed \"largest\" prime $p$.\n- If $N$ is composite, it must have a prime factor, say $g$. Since dividing $N$ by any prime less than or equal to $p$ leaves a remainder of $1$, this prime factor $g$ cannot be in our list. Thus, we must have $g > p$.\n\nIn both cases, we have shown the existence of a prime number greater than $p$.\n\n*This step earns **A1** for deducing that $N$ is either prime or has a prime factor greater than $p$.*"},{"type":"text","order":4,"title":"Conclude the proof","content":"This directly **contradicts** our initial assumption that $p$ was the largest prime number (and that the list of primes was finite).\n\nSince our assumption leads to a contradiction, the assumption must be false.\n\nTherefore, there are **infinitely many** prime numbers.\n\n*This step earns the final **A1** for stating the contradiction and **A1** for concluding that there are infinitely many prime numbers.*"}]}],"generatedAt":"2026-06-09T03:48:34.214Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"28f0d98f-3363-4b89-8b16-46c25504862b:euclid_product","label":"Euclid's Product Construction","steps":[{"type":"text","order":0,"title":"State the assumption","content":"To prove this by contradiction, we start by assuming the opposite of what we want to prove.\n\nAssume that there is a **finite** number of prime numbers. We can write them all down as a complete list:\n\n$$p_1, p_2, p_3, \\dots, p_n$$\n\nwhere $p_n$ is the largest prime number in existence.\n\n*This step earns the first mark ($B1$) for setting up the correct assumption for contradiction.*"},{"type":"text","order":1,"title":"Construct a new integer","content":"Using Euclid's product construction, we define a new integer $N$ by multiplying all of the prime numbers in our list together and then adding $1$:\n\n$$N = (p_1 \\times p_2 \\times p_3 \\times \\dots \\times p_n) + 1$$\n\n*This earns the second mark ($M1$) for constructing the test number $N$.*"},{"type":"text","order":2,"title":"Analyze divisibility","content":"Let's check if any of the prime numbers from our assumed list can divide $N$ exactly.\n\nIf we divide $N$ by any prime $p_i$ from our list, we get:\n\n$$\\frac{N}{p_i} = \\left(\\frac{p_1 \\times p_2 \\times \\dots \\times p_n}{p_i}\\right) + \\frac{1}{p_i}$$\n\nSince $p_i$ is a factor in the numerator of the first term, that division results in an integer. However, the term $\\frac{1}{p_i}$ remains.\n\nTherefore, dividing $N$ by any of the primes $p_1, \\dots, p_n$ leaves a **remainder of $1$**.\n\n*This earns the third mark ($M1$) for establishing the remainder of 1.*"},{"type":"text","order":3,"title":"Deduce prime factor properties","content":"Every integer greater than $1$ must either be a prime number itself or be divisible by a prime number.\n\nSince none of the primes in our finite list $p_1, \\dots, p_n$ can divide $N$ without a remainder, we are left with two possibilities:\n1. $N$ is prime itself (and because $N$ is greater than any $p_i$, it is a new prime not in our list).\n2. $N$ has a prime factor that is not in our original list.\n\n*This earns the fourth mark ($A1$) for deducing that $N$ must either be prime or have a new prime factor.*"},{"type":"text","order":4,"title":"Identify the contradiction","content":"Both possibilities show that there exists a prime number that was not included in our initial list of primes.\n\nThis directly **contradicts** our initial assumption that our list $p_1, p_2, \\dots, p_n$ contained all prime numbers.\n\n*This earns the fifth mark ($A1$) for identifying the contradiction.*"},{"type":"text","order":5,"title":"State the final conclusion","content":"Since our assumption of a finite number of primes leads to a logical contradiction, the assumption must be false.\n\nTherefore, there are **infinitely many prime numbers**.\n\n*This earns the final mark ($A1$) for the complete conclusion.*"}]},{"id":"28f0d98f-3363-4b89-8b16-46c25504862b:factorial_construction","label":"Factorial Construction","steps":[{"type":"text","order":0,"title":"Assume finite primes","content":"To prove the statement by contradiction, we start by assuming the opposite of what we want to prove.\n\nAssume that the set of prime numbers is **finite**. Let this complete list of prime numbers be represented as:\n\n$$\\{p_1, p_2, p_3, \\dots, p\\}$$\n\nwhere $p$ is the **largest prime number** in existence.\n\n*This step earns **B1** for setting up the proof by contradiction with a finite list of primes.*"},{"type":"text","order":1,"title":"Construct the test number","content":"Using our assumption, we construct a new integer, $N$, using the factorial of our largest prime $p$:\n\n$$N = p! + 1$$\n\nRecall that the factorial of $p$ is the product of all positive integers from $1$ up to $p$:\n\n$$p! = p \\times (p - 1) \\times (p - 2) \\times \\dots \\times 2 \\times 1$$\n\n*This step earns **M1** for constructing the value $N = p! + 1$.*"},{"type":"text","order":2,"title":"Analyze divisibility","content":"Let's consider what happens when we divide $N$ by any prime number $q$ from our assumed finite list.\n\nSince $p$ is the largest prime in our list, any prime $q$ must satisfy $q \\le p$. Therefore, $q$ is a factor of $p!$:\n\n$$p! = q \\times K$$ \n\nfor some integer $K$. Substitute this back into our expression for $N$:\n\n$$N = qK + 1$$\n\nDividing $N$ by $q$ gives:\n\n$$\\frac{N}{q} = K + \\frac{1}{q}$$\n\nThis shows that dividing $N$ by any of the primes in our finite list always leaves a **remainder of 1**.\n\n*This step earns **M1** for showing that dividing $N$ by any of the assumed primes leaves a remainder of 1.*"},{"type":"text","order":3,"title":"Find the contradiction","content":"By the Fundamental Theorem of Arithmetic, the integer $N > 1$ must either be prime itself or have a prime factor.\n\n- If $N$ is prime, then since $N = p! + 1 > p$, we have found a prime number greater than our supposed \"largest\" prime $p$.\n- If $N$ is composite, it must have a prime factor, say $g$. Since dividing $N$ by any prime less than or equal to $p$ leaves a remainder of $1$, this prime factor $g$ cannot be in our list. Thus, we must have $g > p$.\n\nIn both cases, we have shown the existence of a prime number greater than $p$.\n\n*This step earns **A1** for deducing that $N$ is either prime or has a prime factor greater than $p$.*"},{"type":"text","order":4,"title":"Conclude the proof","content":"This directly **contradicts** our initial assumption that $p$ was the largest prime number (and that the list of primes was finite).\n\nSince our assumption leads to a contradiction, the assumption must be false.\n\nTherefore, there are **infinitely many** prime numbers.\n\n*This step earns the final **A1** for stating the contradiction and **A1** for concluding that there are infinitely many prime 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1.1 Proof by Contradiction

Question 8

1.1 Proof by Contradiction Questions

1.1 Proof by Contradiction

Question 8

1.1 Proof by Contradiction Questions