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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/proof-6/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":17,"initialQuestionWindow":[{"id":"fe9a1d93-a8a9-4e0d-aca3-70ae3b951191","revisionId":"7803b325-4a93-4e0d-bc4a-5f00793c42d8","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"hard","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"a6e52348-6ec1-465b-bf27-4ebc042ef84c","label":"a","content":"Prove by contradiction that $\\sqrt{3}$ is irrational.","markscheme":"- Assume $\\sqrt{3}$ is rational, so $\\sqrt{3} = \\frac{a}{b}$ where $a, b$ are integers with no common factors. B1\n- Square both sides and rearrange: $3 = \\frac{a^2}{b^2} \\Rightarrow 3b^2 = a^2$ M1\n- Deduce $a^2$ is a multiple of 3, which implies $a$ is a multiple of 3. A1\n- Let $a = 3k$ for some integer $k$, then $3b^2 = (3k)^2 = 9k^2 \\Rightarrow b^2 = 3k^2$ M1\n- Deduce $b^2$ is a multiple of 3, so $b$ is a multiple of 3. A1\n- Since $a$ and $b$ both have a common factor of 3, this contradicts the assumption that they have no common factors. Therefore $\\sqrt{3}$ is irrational. A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Assume $\\sqrt{3} = a/b$ where $a, b$ are integers with no common factors.","code":"B"},{"id":"p2","marks":1,"description":"Square and rearrange to $3b^2 = a^2$.","code":"M"},{"id":"p3","marks":1,"description":"Correctly deduce that $a$ is a multiple of 3.","code":"A"},{"id":"p4","marks":1,"description":"Substitute $a = 3k$ and simplify to $b^2 = 3k^2$.","code":"M"},{"id":"p5","marks":1,"description":"Correctly deduce that $b$ is a multiple of 3.","code":"A"},{"id":"p6","marks":1,"description":"Conclude that the common factor contradicts the initial assumption.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:03:23.615Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"coprime_contradiction","label":"Coprime Fraction Method","steps":[{"type":"text","order":0,"title":"Assume the contrary","content":"To prove that $\\sqrt{3}$ is irrational by contradiction, we begin by assuming that it is actually rational.\n\nBy definition, any rational number can be expressed as a fraction in its simplest form:\n\n$$\\sqrt{3} = \\frac{a}{b}$$\n\nwhere $a$ and $b$ are integers with no common factors (meaning they are **coprime** and $b \\neq 0$).\n\n*(This set up earns the first **B1** mark)*"},{"type":"text","order":1,"title":"Square and rearrange","content":"To eliminate the square root, we square both sides of the equation:\n\n$$3 = \\frac{a^2}{b^2}$$\n\nNow, we multiply both sides by $b^2$ to rearrange the equation into a linear form:\n\n$$3b^2 = a^2$$\n\n*(This algebraic rearrangement earns the **M1** mark)*"},{"type":"text","order":2,"title":"Analyze the variable a","content":"Since $3b^2 = a^2$ and $b$ is an integer, it is clear that $a^2$ must be a multiple of $3$.\n\nBecause $3$ is a prime number, if the square of an integer is divisible by $3$, then the integer itself must also be divisible by $3$. Therefore, **$a$ is a multiple of $3$**.\n\n*(Deducing that $a$ is a multiple of $3$ earns the **A1** mark)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$3b^2 = \\htmlData{anchor=target-a}{\\color{@sky}{a}}^2$"},{"latex":"$$\\text{Since } a \text{ is a multiple of } 3\text{, we let } \\htmlData{anchor=source-a}{\\color{@sky}{a}} = \\color{@amber}{3k} \\text{ for some integer } k$"},{"latex":"$$3b^2 = (\\color{@amber}{3k})^2 = 9k^2$"},{"latex":"$$b^2 = 3k^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"source-a"},{"color":"sky","style":"text-color","anchor":"target-a"}],"connections":[{"to":"target-a","from":"source-a","color":"sky","label":"substitute","style":"solid"}]},"order":3,"title":"Substitute to analyze b","description":"Since $a$ is a multiple of $3$, we can write it as $a = 3k$ where $k$ is an integer. Substituting this back into our original rearranged equation allows us to find the properties of $b$."},{"type":"text","order":4,"title":"Deduce b is a multiple","content":"From our simplified equation:\n\n$$b^2 = 3k^2$$\n\nWe can see that $b^2$ is a multiple of $3$.\n\nFollowing the same logic as before, since $3$ is a prime number and $b^2$ is divisible by $3$, it follows that **$b$ must also be a multiple of $3$**.\n\n*(Setting $a = 3k$ and simplifying earns the **M1** mark, and deducing $b$ is a multiple of $3$ earns the **A1** mark)*"},{"type":"text","order":5,"title":"Identify the contradiction","content":"We have shown that:\n- $a$ is a multiple of $3$\n- $b$ is a multiple of $3$\n\nThis means that both $a$ and $b$ share a common factor of $3$. However, this directly contradicts our initial assumption that $a$ and $b$ are coprime and share no common factors.\n\nSince our original assumption has led to a contradiction, it must be false. \n\nTherefore, **$\\sqrt{3}$ is irrational**.\n\n*(This concluding contradiction argument earns the final **A1** mark)*"}]},{"id":"prime_factorization","label":"Prime Factorization Method","steps":[{"type":"text","order":0,"title":"Set up the contradiction","content":"To prove that $\\sqrt{3}$ is irrational by contradiction, we begin by assuming the opposite: that $\\sqrt{3}$ is rational.\n\nIf $\\sqrt{3}$ is rational, it can be written as the ratio of two positive integers, $a$ and $b$:\n$$\\sqrt{3} = \\frac{a}{b} \\quad \\text{where } a, b \\in \\mathbb{Z}^+$$\n\nNow, square both sides to eliminate the square root:\n$$3 = \\frac{a^2}{b^2}$$\n\nMultiply both sides by $b^2$ to get a relation in terms of integers:\n$$3b^2 = a^2$$\n\n*This initial setup and rearrangement correspond to the **B1** and **M1** marks on typical exam markschemes.*","highlights":[{"text":"Prove by contradiction that $\\sqrt{3}$ is irrational.","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the right side","content":"By the **Fundamental Theorem of Arithmetic**, every positive integer greater than 1 has a unique prime factorization. Let's analyze the multiplicity (the exponent) of the prime factor $3$ on both sides of our equation.\n\nLet the exponent of $3$ in the prime factorization of $a$ be $p$ (where $p \\ge 0$). \n\nWhen we square $a$, the exponents in its prime factorization are doubled. Thus, the exponent of the prime factor $3$ in the factorization of $a^2$ is:\n$$\\text{Exponent of } 3 \\text{ in } a^2 = 2p$$\n\nSince $2p$ is a multiple of $2$, the number of times $3$ appears in the prime factorization of $a^2$ must be **even**."},{"type":"text","order":2,"title":"Analyze the left side","content":"Now, let's look at the left-hand side of the equation, $3b^2$.\n\nLet the exponent of $3$ in the prime factorization of $b$ be $q$ (where $q \\ge 0$).\n\nWhen we square $b$, the exponent of $3$ in $b^2$ is doubled to $2q$ (an even number).\n\nWhen we multiply $b^2$ by $3$, we add exactly one more factor of $3$ to the prime factorization. Therefore, the total exponent of $3$ in the prime factorization of $3b^2$ is:\n$$\\text{Exponent of } 3 \\text{ in } 3b^2 = 2q + 1$$\n\nSince $2q + 1$ is of the form $2k + 1$, the number of times $3$ appears in the prime factorization of $3b^2$ must be **odd**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$3b^2 = a^2$$"},{"latex":"$$$\\text{Exponent of } 3 \\text{ on LHS} = \\htmlData{anchor=lhs-exp}{\\color{@pink}{2q + 1}}$$"},{"latex":"$$$\\text{Exponent of } 3 \\text{ on RHS} = \\htmlData{anchor=rhs-exp}{\\color{@sky}{2p}}$$"}],"highlights":[{"color":"pink","style":"box","anchor":"lhs-exp"},{"color":"sky","style":"box","anchor":"rhs-exp"}],"connections":[{"to":"rhs-exp","from":"lhs-exp","color":"purple","label":"must be equal","style":"dashed"}]},"order":3,"title":"Find the contradiction","description":"For the equation $3b^2 = a^2$ to hold, the prime factorizations of both sides must be identical. This requires the multiplicity of the prime factor $3$ to be equal on both sides: \n\n$$2q + 1 = 2p$$\n\nHowever, $2q + 1$ is an **odd** integer and $2p$ is an **even** integer. An odd integer can never equal an even integer. \n\nThis contradiction proves that our initial assumption—that $\\sqrt{3}$ is rational—must be false. Therefore, $\\sqrt{3}$ is irrational."}]}],"generatedAt":"2026-06-09T11:50:33.540Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"a6e52348-6ec1-465b-bf27-4ebc042ef84c:coprime_contradiction","label":"Coprime Fraction Method","steps":[{"type":"text","order":0,"title":"Assume the contrary","content":"To prove that $\\sqrt{3}$ is irrational by contradiction, we begin by assuming that it is actually rational.\n\nBy definition, any rational number can be expressed as a fraction in its simplest form:\n\n$$\\sqrt{3} = \\frac{a}{b}$$\n\nwhere $a$ and $b$ are integers with no common factors (meaning they are **coprime** and $b \\neq 0$).\n\n*(This set up earns the first **B1** mark)*"},{"type":"text","order":1,"title":"Square and rearrange","content":"To eliminate the square root, we square both sides of the equation:\n\n$$3 = \\frac{a^2}{b^2}$$\n\nNow, we multiply both sides by $b^2$ to rearrange the equation into a linear form:\n\n$$3b^2 = a^2$$\n\n*(This algebraic rearrangement earns the **M1** mark)*"},{"type":"text","order":2,"title":"Analyze the variable a","content":"Since $3b^2 = a^2$ and $b$ is an integer, it is clear that $a^2$ must be a multiple of $3$.\n\nBecause $3$ is a prime number, if the square of an integer is divisible by $3$, then the integer itself must also be divisible by $3$. Therefore, **$a$ is a multiple of $3$**.\n\n*(Deducing that $a$ is a multiple of $3$ earns the **A1** mark)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$3b^2 = \\htmlData{anchor=target-a}{\\color{@sky}{a}}^2$"},{"latex":"$$\\text{Since } a \\text{ is a multiple of } 3\\text{, we let } \\htmlData{anchor=source-a}{\\color{@sky}{a}} = \\color{@amber}{3k} \\text{ for some integer } k$"},{"latex":"$$3b^2 = (\\color{@amber}{3k})^2 = 9k^2$"},{"latex":"$$b^2 = 3k^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"source-a"},{"color":"sky","style":"text-color","anchor":"target-a"}],"connections":[{"to":"target-a","from":"source-a","color":"sky","label":"substitute","style":"solid"}]},"order":3,"title":"Substitute to analyze b","description":"Since $a$ is a multiple of $3$, we can write it as $a = 3k$ where $k$ is an integer. Substituting this back into our original rearranged equation allows us to find the properties of $b$."},{"type":"text","order":4,"title":"Deduce b is a multiple","content":"From our simplified equation:\n\n$$b^2 = 3k^2$$\n\nWe can see that $b^2$ is a multiple of $3$.\n\nFollowing the same logic as before, since $3$ is a prime number and $b^2$ is divisible by $3$, it follows that **$b$ must also be a multiple of $3$**.\n\n*(Setting $a = 3k$ and simplifying earns the **M1** mark, and deducing $b$ is a multiple of $3$ earns the **A1** mark)*"},{"type":"text","order":5,"title":"Identify the contradiction","content":"We have shown that:\n- $a$ is a multiple of $3$\n- $b$ is a multiple of $3$\n\nThis means that both $a$ and $b$ share a common factor of $3$. However, this directly contradicts our initial assumption that $a$ and $b$ are coprime and share no common factors.\n\nSince our original assumption has led to a contradiction, it must be false. \n\nTherefore, **$\\sqrt{3}$ is irrational**.\n\n*(This concluding contradiction argument earns the final **A1** mark)*"}]},{"id":"a6e52348-6ec1-465b-bf27-4ebc042ef84c:prime_factorization","label":"Prime Factorization Method","steps":[{"type":"text","order":0,"title":"Set up the contradiction","content":"To prove that $\\sqrt{3}$ is irrational by contradiction, we begin by assuming the opposite: that $\\sqrt{3}$ is rational.\n\nIf $\\sqrt{3}$ is rational, it can be written as the ratio of two positive integers, $a$ and $b$:\n$$\\sqrt{3} = \\frac{a}{b} \\quad \\text{where } a, b \\in \\mathbb{Z}^+$$\n\nNow, square both sides to eliminate the square root:\n$$3 = \\frac{a^2}{b^2}$$\n\nMultiply both sides by $b^2$ to get a relation in terms of integers:\n$$3b^2 = a^2$$\n\n*This initial setup and rearrangement correspond to the **B1** and **M1** marks on typical exam markschemes.*","highlights":[{"text":"Prove by contradiction that $\\sqrt{3}$ is irrational.","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the right side","content":"By the **Fundamental Theorem of Arithmetic**, every positive integer greater than 1 has a unique prime factorization. Let's analyze the multiplicity (the exponent) of the prime factor $3$ on both sides of our equation.\n\nLet the exponent of $3$ in the prime factorization of $a$ be $p$ (where $p \\ge 0$). \n\nWhen we square $a$, the exponents in its prime factorization are doubled. Thus, the exponent of the prime factor $3$ in the factorization of $a^2$ is:\n$$\\text{Exponent of } 3 \\text{ in } a^2 = 2p$$\n\nSince $2p$ is a multiple of $2$, the number of times $3$ appears in the prime factorization of $a^2$ must be **even**."},{"type":"text","order":2,"title":"Analyze the left side","content":"Now, let's look at the left-hand side of the equation, $3b^2$.\n\nLet the exponent of $3$ in the prime factorization of $b$ be $q$ (where $q \\ge 0$).\n\nWhen we square $b$, the exponent of $3$ in $b^2$ is doubled to $2q$ (an even number).\n\nWhen we multiply $b^2$ by $3$, we add exactly one more factor of $3$ to the prime factorization. Therefore, the total exponent of $3$ in the prime factorization of $3b^2$ is:\n$$\\text{Exponent of } 3 \\text{ in } 3b^2 = 2q + 1$$\n\nSince $2q + 1$ is of the form $2k + 1$, the number of times $3$ appears in the prime factorization of $3b^2$ must be **odd**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$3b^2 = a^2$$"},{"latex":"$$$\\text{Exponent of } 3 \\text{ on LHS} = \\htmlData{anchor=lhs-exp}{\\color{@pink}{2q + 1}}$$"},{"latex":"$$$\\text{Exponent of } 3 \\text{ on RHS} = \\htmlData{anchor=rhs-exp}{\\color{@sky}{2p}}$$"}],"highlights":[{"color":"pink","style":"box","anchor":"lhs-exp"},{"color":"sky","style":"box","anchor":"rhs-exp"}],"connections":[{"to":"rhs-exp","from":"lhs-exp","color":"purple","label":"must be equal","style":"dashed"}]},"order":3,"title":"Find the contradiction","description":"For the equation $3b^2 = a^2$ to hold, the prime factorizations of both sides must be identical. This requires the multiplicity of the prime factor $3$ to be equal on both sides: \n\n$$2q + 1 = 2p$$\n\nHowever, $2q + 1$ is an **odd** integer and $2p$ is an **even** integer. An odd integer can never equal an even integer. \n\nThis contradiction proves that our initial assumption—that $\\sqrt{3}$ is rational—must be false. Therefore, $\\sqrt{3}$ is irrational."}]}],"generatedAt":"2026-06-09T11:50:33.613Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":72},"walkthroughVideo":null},{"id":"6c4ee647-e514-46f6-88b7-ec0733bbe1ed","revisionId":"868fe90c-9d56-45ad-849c-8097c84eeb25","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"hard","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"06518a2f-569b-4542-8bc1-666bf1872fb0","label":"a","content":"Using the method of proof by contradiction, prove that there is no greatest prime number.","markscheme":"- Assume that there is a greatest prime number, $L$. B1\n- Consider the product of all prime numbers from 2 up to $L$: $X = 2 \\times 3 \\times 5 \\times \\dots \\times L$. M1\n- Define $Y = X + 1$. M1\n- Argue that $Y$ is not divisible by any prime $p$ such that $p \\le L$, as division by any such prime leaves a remainder of 1. A1\n- Deduce that $Y$ is either a prime number itself or it has a prime factor that is greater than $L$. M1\n- Conclude that this contradicts the assumption that $L$ is the greatest prime number, therefore no greatest prime exists. A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Assume that a largest prime number $L$ exists.","code":"B"},{"id":"p2","marks":1,"description":"Form the product of all primes up to $L$.","code":"M"},{"id":"p3","marks":1,"description":"Form a new number by adding 1 to that product.","code":"M"},{"id":"p4","marks":1,"description":"Demonstrate that no prime up to $L$ is a factor of the new number.","code":"A"},{"id":"p5","marks":1,"description":"Establish that a prime exists that is not in the original list.","code":"M"},{"id":"p6","marks":1,"description":"Conclude that the existence of a greatest prime is impossible.","code":"A"}],"finalAnswers":["There is no greatest prime number"],"generatedAt":"2026-05-06T14:05:19.695Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"euclid_product","label":"Euclid's Product Construction","steps":[{"type":"text","order":0,"title":"Assume the opposite","content":"To prove this statement by contradiction, we begin by assuming the opposite of what we want to show.\n\nAssume that there is a finite number of primes, meaning there exists a **greatest prime number**, which we will call $L$.\n\nThus, our complete list of all prime numbers is:\n$$2, 3, 5, 7, \\dots, L$$"},{"type":"text","order":1,"title":"Construct the product","content":"Next, we construct a new integer, $Y$, by multiplying all of these prime numbers together and adding $1$.\n\nLet $X$ be the product of all prime numbers up to $L$:\n$$X = 2 \\times 3 \\times 5 \\times \\dots \\times L$$\n\nNow, we define $Y$ as:\n$$Y = X + 1 = (2 \\times 3 \\times 5 \\times \\dots \\times L) + 1$$"},{"type":"text","order":2,"title":"Test divisibility of Y","content":"Let's consider what happens when we divide $Y$ by any of the prime numbers, $p$, from our list ($2 \\le p \\le L$).\n\nSince $p$ is a factor of the product $X$, dividing $X$ by $p$ leaves no remainder. \n\nTherefore, dividing $Y = X + 1$ by $p$ will always leave a remainder of $1$:\n\n$$\\frac{Y}{p} = \\frac{X}{p} + \\frac{1}{p}$$\n\nThis means that $Y$ is **not divisible** by any prime $p \\le L$."},{"type":"text","order":3,"title":"Analyze the prime factors","content":"According to the Fundamental Theorem of Arithmetic, every integer greater than $1$ must either be a prime number itself or be divisible by a prime number. Let's analyze these two cases for $Y$:\n\n- **Case 1: $Y$ is prime**\n Since $Y > L$, we have found a prime number greater than $L$.\n\n- **Case 2: $Y$ is composite**\n Since $Y$ is not divisible by any prime up to $L$, any prime factor of $Y$ must be **greater than $L$**."},{"type":"text","order":4,"title":"Reach a contradiction","content":"In both cases, we have shown the existence of a prime number that is greater than $L$.\n\nThis directly contradicts our initial assumption that $L$ is the greatest prime number.\n\nTherefore, our initial assumption must be false. We conclude that **there is no greatest prime number** (meaning the number of primes is infinite)."}]},{"id":"factorial_construction","label":"Factorial Construction","steps":[{"type":"text","order":0,"title":"Assume a greatest prime","content":"To prove there is no greatest prime number by contradiction, we begin by assuming the opposite of what we want to prove.\n\nAssume that there **is** a greatest prime number, which we will call $L$.\n\nThis means that the set of all prime numbers is finite, and any prime number $p$ must satisfy $p \\le L$.\n\n*(This earns the first mark **B1**)*"},{"type":"text","order":1,"title":"Construct the factorial expression","content":"Consider the factorial of $L$, which is the product of all integers from 1 up to $L$:\n\n$$L! = L \\times (L-1) \\times (L-2) \\times \\dots \\times 2 \\times 1$$\n\nNow, let us define a new integer $N$ as:\n\n$$N = L! + 1$$\n\n*(Constructing this expression earns the **M1** marks for creating a product of terms plus one)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=ndef}{N} = \\htmlData{anchor=lfact}{L!} + 1$"},{"text":"Let $p$ be any prime number such that $p \\le L$. Since $p$ is an integer in the range $1$ to $L$, it is a factor of $L!$."},{"latex":"$$\\frac{\\htmlData{anchor=ndiv}{N}}{\\htmlData{anchor=pdiv1}{\\color{@sky}{p}}} = \\frac{\\htmlData{anchor=ldiv}{L!}}{\\htmlData{anchor=pdiv2}{\\color{@sky}{p}}} + \\frac{1}{\\color{@sky}{p}}$"},{"text":"Since $p$ divides $L!$, the term $\\frac{L!}{p}$ is an integer, leaving a remainder of $1$ when $N$ is divided by $p$."}],"highlights":[{"color":"sky","style":"text-color","anchor":"pdiv1"},{"color":"sky","style":"text-color","anchor":"pdiv2"}],"connections":[{"to":"ndiv","from":"ndef","color":"purple","label":"substitute","style":"solid"},{"to":"ldiv","from":"lfact","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Analyze divisibility by primes","description":"We divide the entire expression for $N$ by a prime $p \\le L$ to inspect its remainder. Since the remainder is 1, $N$ is not divisible by any prime $p \\le L$. (This earns the **A1** mark)"},{"type":"text","order":3,"title":"Deduce a larger prime","content":"According to the Fundamental Theorem of Arithmetic, every integer greater than 1 is either a prime number itself or can be factored into prime numbers.\n\nSince $N > 1$, it must have at least one prime factor, let us call it $q$.\n\nFrom our previous step, we established that **no** prime $p$ less than or equal to $L$ can divide $N$. Therefore, the prime factor $q$ cannot be any of the primes less than or equal to $L$.\n\nThus, we must have:\n\n$$q > L$$\n\n*(This deduction earns **M1**)*"},{"type":"text","order":4,"title":"State the contradiction","content":"We have shown that $N$ must have a prime factor $q$ such that $q > L$.\n\nThis directly **contradicts** our initial assumption that $L$ is the greatest prime number.\n\nTherefore, our assumption must be false. We conclude that there is no greatest prime number.\n\n*(This final conclusion earns **A1**)*"}]}],"generatedAt":"2026-06-09T06:39:31.170Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"06518a2f-569b-4542-8bc1-666bf1872fb0:euclid_product","label":"Euclid's Product Construction","steps":[{"type":"text","order":0,"title":"Assume the opposite","content":"To prove this statement by contradiction, we begin by assuming the opposite of what we want to show.\n\nAssume that there is a finite number of primes, meaning there exists a **greatest prime number**, which we will call $L$.\n\nThus, our complete list of all prime numbers is:\n$$2, 3, 5, 7, \\dots, L$$"},{"type":"text","order":1,"title":"Construct the product","content":"Next, we construct a new integer, $Y$, by multiplying all of these prime numbers together and adding $1$.\n\nLet $X$ be the product of all prime numbers up to $L$:\n$$X = 2 \\times 3 \\times 5 \\times \\dots \\times L$$\n\nNow, we define $Y$ as:\n$$Y = X + 1 = (2 \\times 3 \\times 5 \\times \\dots \\times L) + 1$$"},{"type":"text","order":2,"title":"Test divisibility of Y","content":"Let's consider what happens when we divide $Y$ by any of the prime numbers, $p$, from our list ($2 \\le p \\le L$).\n\nSince $p$ is a factor of the product $X$, dividing $X$ by $p$ leaves no remainder. \n\nTherefore, dividing $Y = X + 1$ by $p$ will always leave a remainder of $1$:\n\n$$\\frac{Y}{p} = \\frac{X}{p} + \\frac{1}{p}$$\n\nThis means that $Y$ is **not divisible** by any prime $p \\le L$."},{"type":"text","order":3,"title":"Analyze the prime factors","content":"According to the Fundamental Theorem of Arithmetic, every integer greater than $1$ must either be a prime number itself or be divisible by a prime number. Let's analyze these two cases for $Y$:\n\n- **Case 1: $Y$ is prime**\n Since $Y > L$, we have found a prime number greater than $L$.\n\n- **Case 2: $Y$ is composite**\n Since $Y$ is not divisible by any prime up to $L$, any prime factor of $Y$ must be **greater than $L$**."},{"type":"text","order":4,"title":"Reach a contradiction","content":"In both cases, we have shown the existence of a prime number that is greater than $L$.\n\nThis directly contradicts our initial assumption that $L$ is the greatest prime number.\n\nTherefore, our initial assumption must be false. We conclude that **there is no greatest prime number** (meaning the number of primes is infinite)."}]},{"id":"06518a2f-569b-4542-8bc1-666bf1872fb0:factorial_construction","label":"Factorial Construction","steps":[{"type":"text","order":0,"title":"Assume a greatest prime","content":"To prove there is no greatest prime number by contradiction, we begin by assuming the opposite of what we want to prove.\n\nAssume that there **is** a greatest prime number, which we will call $L$.\n\nThis means that the set of all prime numbers is finite, and any prime number $p$ must satisfy $p \\le L$.\n\n*(This earns the first mark **B1**)*"},{"type":"text","order":1,"title":"Construct the factorial expression","content":"Consider the factorial of $L$, which is the product of all integers from 1 up to $L$:\n\n$$L! = L \\times (L-1) \\times (L-2) \\times \\dots \\times 2 \\times 1$$\n\nNow, let us define a new integer $N$ as:\n\n$$N = L! + 1$$\n\n*(Constructing this expression earns the **M1** marks for creating a product of terms plus one)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=ndef}{N} = \\htmlData{anchor=lfact}{L!} + 1$"},{"text":"Let $p$ be any prime number such that $p \\le L$. Since $p$ is an integer in the range $1$ to $L$, it is a factor of $L!$."},{"latex":"$$\\frac{\\htmlData{anchor=ndiv}{N}}{\\htmlData{anchor=pdiv1}{\\color{@sky}{p}}} = \\frac{\\htmlData{anchor=ldiv}{L!}}{\\htmlData{anchor=pdiv2}{\\color{@sky}{p}}} + \\frac{1}{\\color{@sky}{p}}$"},{"text":"Since $p$ divides $L!$, the term $\\frac{L!}{p}$ is an integer, leaving a remainder of $1$ when $N$ is divided by $p$."}],"highlights":[{"color":"sky","style":"text-color","anchor":"pdiv1"},{"color":"sky","style":"text-color","anchor":"pdiv2"}],"connections":[{"to":"ndiv","from":"ndef","color":"purple","label":"substitute","style":"solid"},{"to":"ldiv","from":"lfact","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Analyze divisibility by primes","description":"We divide the entire expression for $N$ by a prime $p \\le L$ to inspect its remainder. Since the remainder is 1, $N$ is not divisible by any prime $p \\le L$. (This earns the **A1** mark)"},{"type":"text","order":3,"title":"Deduce a larger prime","content":"According to the Fundamental Theorem of Arithmetic, every integer greater than 1 is either a prime number itself or can be factored into prime numbers.\n\nSince $N > 1$, it must have at least one prime factor, let us call it $q$.\n\nFrom our previous step, we established that **no** prime $p$ less than or equal to $L$ can divide $N$. Therefore, the prime factor $q$ cannot be any of the primes less than or equal to $L$.\n\nThus, we must have:\n\n$$q > L$$\n\n*(This deduction earns **M1**)*"},{"type":"text","order":4,"title":"State the contradiction","content":"We have shown that $N$ must have a prime factor $q$ such that $q > L$.\n\nThis directly **contradicts** our initial assumption that $L$ is the greatest prime number.\n\nTherefore, our assumption must be false. We conclude that there is no greatest prime number.\n\n*(This final conclusion earns **A1**)*"}]}],"generatedAt":"2026-06-09T06:39:41.100Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":48},"walkthroughVideo":null},{"id":"c7d9a82a-fa0f-4093-aeb3-1480c2a1ac5f","revisionId":"7e83398a-468f-44eb-9f97-1c2b7256a788","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"hard","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"83f81360-02d8-4603-ba8f-376d1cfd2617","label":"a","content":"Prove by contradiction that the product of a non-zero rational number and an irrational number is irrational.","markscheme":"- Assume the product is rational: $r \\times i = q$, where $r = \\frac{a}{b}$ ($a, b \\neq 0$) and $q = \\frac{c}{d}$ are rational. B1\n- Rearrange to isolate the irrational number: $i = \\frac{c}{d} \\div \\frac{a}{b}$ M1\n- Simplify to a single fraction using the reciprocal: $i = \\frac{bc}{ad}$ M1\n- State that as $a, b, c, d$ are integers and $ad \\neq 0$, then $bc$ and $ad$ are integers. A1\n- State that this implies $i$ is rational, which is a contradiction. A1\n- Conclude that the product of a non-zero rational and an irrational number must be irrational. A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Assume the product is rational and define rational numbers as ratios of integers.","code":"B"},{"id":"p2","marks":1,"description":"Isolate the irrational number using division.","code":"M"},{"id":"p3","marks":1,"description":"Combine into a single fraction $bc/ad$.","code":"M"},{"id":"p4","marks":1,"description":"State that $bc$ and $ad$ are integers.","code":"A"},{"id":"p5","marks":1,"description":"Identify the contradiction that an irrational number is equal to a rational number.","code":"A"},{"id":"p6","marks":1,"description":"Final concluding statement.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:03:35.615Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"fraction_definition","label":"Using Fraction Definitions","steps":[{"type":"text","order":0,"title":"Set up the contradiction","content":"To prove the statement by contradiction, we start by assuming the opposite of what we want to prove. Let $r$ be a non-zero rational number and $i$ be an irrational number. We assume that their product, which we will call $p$, is a rational number:\n\n$$p = r \\times i$$"},{"type":"text","order":1,"title":"Define rational numbers as fractions","content":"By definition, any rational number can be expressed as a fraction of two integers. Since we assumed $r$ and $p$ are rational:\n\n1. Let $r = \\frac{a}{b}$ where $a$ and $b$ are integers. Since $r$ is non-zero, we have $a \\neq 0$ and $b \\neq 0$.\n2. Let $p = \\frac{c}{d}$ where $c$ and $d$ are integers and $d \\neq 0$.\n\nSubstituting these fraction definitions into our product equation gives:\n\n$$\\frac{a}{b} \\times i = \\frac{c}{d}$$"},{"type":"text","order":2,"title":"Isolate the irrational number","content":"To isolate the irrational number $i$, we divide both sides by the non-zero fraction $\\frac{a}{b}$:\n\n$$i = \\frac{c}{d} \\div \\frac{a}{b}$$\n\nTo divide by a fraction, we multiply by its reciprocal:\n\n$$i = \\frac{c}{d} \\times \\frac{b}{a}$$\n\n$$i = \\frac{bc}{ad}$$"},{"type":"text","order":3,"title":"Identify the contradiction","content":"Now we analyze the properties of our numerator and denominator:\n\n- The numerator $bc$ is the product of two integers ($b$ and $c$), so $bc$ is also an integer.\n- The denominator $ad$ is the product of two non-zero integers ($a$ and $d$), so $ad$ is a non-zero integer ($ad \\neq 0$).\n\nSince $i = \\frac{bc}{ad}$ is the ratio of two integers with a non-zero denominator, $i$ must be a rational number by definition."},{"type":"text","order":4,"title":"Conclude the proof","content":"Our deduction that $i$ is rational directly contradicts our initial given condition that $i$ is an irrational number. \n\nTherefore, our initial assumption that the product is rational must be false. We conclude that the product of a non-zero rational number and an irrational number is always irrational."}]},{"id":"closure_properties","label":"Using Rational Closure Properties","steps":[{"type":"text","order":0,"title":"Set up the assumption","content":"To prove this by contradiction, we begin by assuming the opposite of what we want to prove.\n\nLet's assume that the product of a non-zero rational number $r$ and an irrational number $i$ is a rational number $q$:\n\n$$r \\times i = q$$\n\nSince $r$ and $q$ are rational numbers, we can express them as fractions of integers:\n- $r = \\frac{a}{b}$, where $a, b \\in \\mathbb{Z}$ and $a, b \\neq 0$ (since $r$ is non-zero).\n- $q = \\frac{c}{d}$, where $c, d \\in \\mathbb{Z}$ and $d \\neq 0$.\n\n*This initial setup and definition of terms earns the first mark **[B1]**.*","highlights":[{"text":"product of a non-zero rational number and an irrational number is irrational","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=eq1}{r \\times i = q} \\implies i = \\frac{\\htmlData{anchor=q-val}{\\color{@purple}{q}}}{\\htmlData{anchor=r-val}{\\color{@sky}{r}}}$"},{"latex":"$$i = \\htmlData{anchor=q-frac}{\\color{@purple}{\\frac{c}{d}}} \\div \\htmlData{anchor=r-frac}{\\color{@sky}{\\frac{a}{b}}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"q-val"},{"color":"sky","style":"text-color","anchor":"r-val"},{"color":"purple","style":"text-color","anchor":"q-frac"},{"color":"sky","style":"text-color","anchor":"r-frac"}],"connections":[{"to":"q-frac","from":"q-val","color":"purple","label":"substitute q","style":"solid"},{"to":"r-frac","from":"r-val","color":"sky","label":"substitute r","style":"solid"}]},"order":1,"title":"Isolate the irrational number","description":"We rearrange the product equation to isolate the irrational number $i$. Since $r$ is non-zero, we can divide both sides by $r$. This rearrangement step is worth the second mark **[M1]**."},{"type":"text","order":2,"title":"Simplify to a single fraction","content":"To simplify the division of fractions, we multiply by the reciprocal of the divisor:\n\n$$i = \\frac{c}{d} \\times \\frac{b}{a}$$\n\nCombining this into a single fraction gives us:\n\n$$i = \\frac{bc}{ad}$$\n\n*This algebraic simplification to a single fraction earns the third mark **[M1]**.*"},{"type":"text","order":3,"title":"Apply closure properties of integers","content":"Let's look at the numerator and the denominator of our simplified fraction:\n- Since $b, c \\in \\mathbb{Z}$, their product $bc$ must be an integer (integers are closed under multiplication).\n- Since $a, d \\in \\mathbb{Z}$, their product $ad$ must also be an integer.\n- Since $a \\neq 0$ and $d \\neq 0$, the product $ad \\neq 0$.\n\n*Explaining why the numerator and denominator are non-zero integers earns the fourth mark **[A1]**.*"},{"type":"text","order":4,"title":"State the contradiction","content":"Since $bc$ and $ad$ are integers and $ad \\neq 0$, the expression $\\frac{bc}{ad}$ fits the definition of a rational number.\n\nThis means that $i$ must be a rational number:\n\n$$i \\in \\mathbb{Q}$$\n\nHowever, this directly contradicts our original statement that $i$ is an **irrational** number.\n\n*Identifying this contradiction earns the fifth mark **[A1]**.*"},{"type":"text","order":5,"title":"Draw the final conclusion","content":"Since our assumption that the product is rational led to a logical contradiction, the assumption must be false.\n\nTherefore, the product of a non-zero rational number and an irrational number must be irrational.\n\n*Stating this final conclusion completes the proof and earns the final mark **[A1]**.*"}]}],"generatedAt":"2026-06-09T09:24:25.842Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"83f81360-02d8-4603-ba8f-376d1cfd2617:fraction_definition","label":"Using Fraction Definitions","steps":[{"type":"text","order":0,"title":"Set up the contradiction","content":"To prove the statement by contradiction, we start by assuming the opposite of what we want to prove. Let $r$ be a non-zero rational number and $i$ be an irrational number. We assume that their product, which we will call $p$, is a rational number:\n\n$$p = r \\times i$$"},{"type":"text","order":1,"title":"Define rational numbers as fractions","content":"By definition, any rational number can be expressed as a fraction of two integers. Since we assumed $r$ and $p$ are rational:\n\n1. Let $r = \\frac{a}{b}$ where $a$ and $b$ are integers. Since $r$ is non-zero, we have $a \\neq 0$ and $b \\neq 0$.\n2. Let $p = \\frac{c}{d}$ where $c$ and $d$ are integers and $d \\neq 0$.\n\nSubstituting these fraction definitions into our product equation gives:\n\n$$\\frac{a}{b} \\times i = \\frac{c}{d}$$"},{"type":"text","order":2,"title":"Isolate the irrational number","content":"To isolate the irrational number $i$, we divide both sides by the non-zero fraction $\\frac{a}{b}$:\n\n$$i = \\frac{c}{d} \\div \\frac{a}{b}$$\n\nTo divide by a fraction, we multiply by its reciprocal:\n\n$$i = \\frac{c}{d} \\times \\frac{b}{a}$$\n\n$$i = \\frac{bc}{ad}$$"},{"type":"text","order":3,"title":"Identify the contradiction","content":"Now we analyze the properties of our numerator and denominator:\n\n- The numerator $bc$ is the product of two integers ($b$ and $c$), so $bc$ is also an integer.\n- The denominator $ad$ is the product of two non-zero integers ($a$ and $d$), so $ad$ is a non-zero integer ($ad \\neq 0$).\n\nSince $i = \\frac{bc}{ad}$ is the ratio of two integers with a non-zero denominator, $i$ must be a rational number by definition."},{"type":"text","order":4,"title":"Conclude the proof","content":"Our deduction that $i$ is rational directly contradicts our initial given condition that $i$ is an irrational number. \n\nTherefore, our initial assumption that the product is rational must be false. We conclude that the product of a non-zero rational number and an irrational number is always irrational."}]},{"id":"83f81360-02d8-4603-ba8f-376d1cfd2617:closure_properties","label":"Using Rational Closure Properties","steps":[{"type":"text","order":0,"title":"Set up the assumption","content":"To prove this by contradiction, we begin by assuming the opposite of what we want to prove.\n\nLet's assume that the product of a non-zero rational number $r$ and an irrational number $i$ is a rational number $q$:\n\n$$r \\times i = q$$\n\nSince $r$ and $q$ are rational numbers, we can express them as fractions of integers:\n- $r = \\frac{a}{b}$, where $a, b \\in \\mathbb{Z}$ and $a, b \\neq 0$ (since $r$ is non-zero).\n- $q = \\frac{c}{d}$, where $c, d \\in \\mathbb{Z}$ and $d \\neq 0$.\n\n*This initial setup and definition of terms earns the first mark **[B1]**.*","highlights":[{"text":"product of a non-zero rational number and an irrational number is irrational","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=eq1}{r \\times i = q} \\implies i = \\frac{\\htmlData{anchor=q-val}{\\color{@purple}{q}}}{\\htmlData{anchor=r-val}{\\color{@sky}{r}}}$"},{"latex":"$$i = \\htmlData{anchor=q-frac}{\\color{@purple}{\\frac{c}{d}}} \\div \\htmlData{anchor=r-frac}{\\color{@sky}{\\frac{a}{b}}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"q-val"},{"color":"sky","style":"text-color","anchor":"r-val"},{"color":"purple","style":"text-color","anchor":"q-frac"},{"color":"sky","style":"text-color","anchor":"r-frac"}],"connections":[{"to":"q-frac","from":"q-val","color":"purple","label":"substitute q","style":"solid"},{"to":"r-frac","from":"r-val","color":"sky","label":"substitute r","style":"solid"}]},"order":1,"title":"Isolate the irrational number","description":"We rearrange the product equation to isolate the irrational number $i$. Since $r$ is non-zero, we can divide both sides by $r$. This rearrangement step is worth the second mark **[M1]**."},{"type":"text","order":2,"title":"Simplify to a single fraction","content":"To simplify the division of fractions, we multiply by the reciprocal of the divisor:\n\n$$i = \\frac{c}{d} \\times \\frac{b}{a}$$\n\nCombining this into a single fraction gives us:\n\n$$i = \\frac{bc}{ad}$$\n\n*This algebraic simplification to a single fraction earns the third mark **[M1]**.*"},{"type":"text","order":3,"title":"Apply closure properties of integers","content":"Let's look at the numerator and the denominator of our simplified fraction:\n- Since $b, c \\in \\mathbb{Z}$, their product $bc$ must be an integer (integers are closed under multiplication).\n- Since $a, d \\in \\mathbb{Z}$, their product $ad$ must also be an integer.\n- Since $a \\neq 0$ and $d \\neq 0$, the product $ad \\neq 0$.\n\n*Explaining why the numerator and denominator are non-zero integers earns the fourth mark **[A1]**.*"},{"type":"text","order":4,"title":"State the contradiction","content":"Since $bc$ and $ad$ are integers and $ad \\neq 0$, the expression $\\frac{bc}{ad}$ fits the definition of a rational number.\n\nThis means that $i$ must be a rational number:\n\n$$i \\in \\mathbb{Q}$$\n\nHowever, this directly contradicts our original statement that $i$ is an **irrational** number.\n\n*Identifying this contradiction earns the fifth mark **[A1]**.*"},{"type":"text","order":5,"title":"Draw the final conclusion","content":"Since our assumption that the product is rational led to a logical contradiction, the assumption must be false.\n\nTherefore, the product of a non-zero rational number and an irrational number must be irrational.\n\n*Stating this final conclusion completes the proof and earns the final mark **[A1]**.*"}]}],"generatedAt":"2026-06-09T09:24:36.321Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__wjec","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-412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Proof (A-level only)","topicId":"23c24b47-7b3c-4108-bb70-3fe34105de38","topicSidebar":"$L3c","topicSlug":"proof-6","topicTitle":"3.1.1 Proof (A-level only)"}]}],"$L3d","$L3e"]}]

3.1.1 Proof (A-level only)

Question 5

3.1.1 Proof (A-level only) Questions

3.1.1 Proof (A-level only)

Question 5

3.1.1 Proof (A-level only) Questions