d":"2bae41ec-5c98-45c7-ad00-d08acf04da70","label":null,"content":"Use the method of proof by contradiction to prove that \$ S \$ is an irrational number.","markscheme":"- Assume $S$ is a rational number, such that $S = \\frac{p}{q}$ where $p, q \\in \\mathbb{Z}$ and $q \\neq 0$ B1\n- Rearrange the equation $\\log_{2} 3 = \\frac{p}{q}$ using laws of logarithms to index form $2^{\\frac{p}{q}} = 3$, and hence $2^p = 3^q$ M1\n- Explain that since $S > 0$, $p$ and $q$ can be assumed to be positive integers; thus $2^p$ is an even number and $3^q$ is an odd number. This is a contradiction, so $S$ must be irrational A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"parity","label":"Parity Argument","steps":[{"type":"text","order":0,"title":"State the contradiction assumption","content":"To prove that $S = \\log_2 3$ is irrational using proof by contradiction, we begin by assuming the opposite. \n\nLet's assume $S$ is a **rational** number. By definition, any rational number can be written as a fraction of two integers. So, we can write:\n\n$$S = \\log_2 3 = \\frac{p}{q}$$\n\nwhere $p$ and $q$ are integers with $q \\neq 0$.\n\nSince $3 > 1$ and the base of the logarithm is $2$, we know that $\\log_2 3 > 0$. Therefore, both $p$ and $q$ can be assumed to be **positive integers** ($p, q \\in \\mathbb{Z}^+$).\n\n*Award yourself the first mark B1 for correctly setting up this assumption.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\log_2 3 = \\frac{\\htmlData{anchor=p-val}{\\color{@sky}{p}}}{\\htmlData{anchor=q-val}{\\color{@amber}{q}}}$"},{"text":"Rewrite this logarithmic equation in its exponential (index) form:"},{"latex":"$$2^{\\frac{\\htmlData{anchor=p-pow}{\\color{@sky}{p}}}{\\htmlData{anchor=q-pow}{\\color{@amber}{q}}}} = 3$"},{"text":"Now, raise both sides of the equation to the power of $q$ to eliminate the fraction in the exponent:"},{"latex":"$$\\left(2^{\\frac{\\color{@sky}{p}}{\\color{@amber}{q}}}\\right)^{\\color{@amber}{q}} = 3^{\\color{@amber}{q}}$"},{"latex":"$$2^{\\htmlData{anchor=p-final}{\\color{@sky}{p}}} = 3^{\\htmlData{anchor=q-final}{\\color{@amber}{q}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"p-val"},{"color":"amber","style":"text-color","anchor":"q-val"}],"connections":[{"to":"p-pow","from":"p-val","color":"sky","label":null,"style":"solid"},{"to":"q-pow","from":"q-val","color":"amber","label":null,"style":"solid"},{"to":"p-final","from":"p-pow","color":"sky","label":null,"style":"solid"},{"to":"q-final","from":"q-pow","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Rearrange to index form","description":"We convert the logarithm to exponential form and eliminate the fraction to get a clean equation with integer exponents."},{"type":"text","order":2,"title":"Analyze parity and find contradiction","content":"Now, let's analyze the properties of both sides of our equation: $$2^p = 3^q$$\n\n- **Left-Hand Side ($2^p$):** Since $p$ is a positive integer ($p \\geq 1$), $2^p$ is a product of 2s, which means it must be an **even** number.\n- **Right-Hand Side ($3^q$):** Since $q$ is a positive integer ($q \\geq 1$), $3^q$ is a product of odd prime factors (3s), which means it must be an **odd** number.\n\nAn even number can never be equal to an odd number:\n\n$$\\text{Even} \\neq \\text{Odd}$$\n\nThis is a contradiction! Our initial assumption that $S$ is rational must be false. Therefore, $S = \\log_2 3$ is an **irrational** number.\n\n*Award yourself M1 for rearranging the logarithmic equation into the form $2^p = 3^q$, and A1 for completing the proof by identifying the even/odd parity contradiction.*"}]},{"id":"primefactorization","label":"Unique Prime Factorization","steps":[{"type":"text","order":0,"title":"Assume S is rational","content":"To prove that $S = \\log_2 3$ is irrational using contradiction, we begin by assuming the opposite. \n\nAssume $S$ is a **rational number**. \n\nSince $3 > 2^0 = 1$, we know that $\\log_2

Algebraic Methods

Question 175

Algebraic Methods Questions

Algebraic Methods

Question 175

Algebraic Methods Questions