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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","tier":null,"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 numbers.*"}]}],"generatedAt":"2026-06-09T03:48:48.306Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__aqa","summaries":[{"id":"fd7ef5e5-e672-484d-827d-216a0bf187eb","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"004f1715-7f20-44bd-b303-beb33cb8f946","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"0f28af03-b719-4ce4-8b8a-9e1d7a79b4a7","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"135c4fa2-edaf-426c-aff5-6dff03228d23","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"26f0bc1d-bd36-470c-847c-ee92ef94d932","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"41bbfce2-2bcf-4f6a-b290-e3abcbb7a077","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"510dbdcb-d107-4e6d-ab01-cf07807fc6f9","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"6587ad34-dbe7-4eaf-93db-d0d216186419","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6b925623-c577-4449-a2dd-7a78f36a658f","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"be93578c-b7d2-4df6-acd7-6ef98a572b2c","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"e56fe79b-a1e4-4b6d-9dca-a4042e1cf70b","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"f5006b00-71ec-4fe1-bfd6-a3689efa8c22","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"fe9a1d93-a8a9-4e0d-aca3-70ae3b951191","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"b5b2f95d-72a6-49b2-9d75-4662197e54af","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"23ef56dd-b092-41bc-abe8-4a2e89a6592d","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"47c36499-dd4e-4e1f-92ce-655791bcc3c1","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6c36ee4d-6b4b-4ff9-accd-9ae7dbccb7ec","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6c4ee647-e514-46f6-88b7-ec0733bbe1ed","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"8dc028ae-d62c-450b-ace3-35c6f6f6a7f8","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"93607e6d-2a9b-443f-a9bb-40c9577b10e1","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"9e651ba3-893d-4d7b-a060-3cd50b14dc4b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"a3db02dd-cb03-45ff-bdb6-a8d761189719","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"aa3791df-7481-43a1-b0b9-cb8e3bcc9c44","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"c7d9a82a-fa0f-4093-aeb3-1480c2a1ac5f","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"d4188522-4341-49ba-a1c1-b0d2cd27b056","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"fe45c078-ebe7-4562-88d0-081c3c872bf2","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"9b74fb76-e5d9-40d0-9ed3-b77c0f17ab33","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"13c9487f-fdfe-413a-bd71-4821d6b1f4c3","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"2228aba7-e076-4386-af3a-881069ea3ae9","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"3f64531b-355c-4f81-9254-e606a290abcd","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"4470672a-f3bb-48d3-96e8-994a0a98913b","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"5198bcf0-d6eb-42d9-8112-78ad2e6eaba6","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6c7892f5-02da-412a-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Structure and methods of proof","topicId":"187b7f27-807a-40aa-bfcb-044c67c27c35","topicSidebar":"$L3d","topicSlug":"structure-and-methods-of-proof","topicTitle":"1.4.1 Structure and methods of proof"}]}],"$L3e","$L3f"]}]

1.4.1 Structure and methods of proof

Question 9

1.4.1 Structure and methods of proof Questions

1.4.1 Structure and methods of proof

Question 9

1.4.1 Structure and methods of proof Questions