IB Maths AA HL Proof Questions

1. Understand the logic that a single case where the hypothesis ($n$ is prime) is true but the conclusion ($2^n-1$ is prime) is false will perfectly disprove the entire statement.

2. Test small prime numbers. For $n=2$, $2^2-1=3$ (prime). For $n=3$, $2^3-1=7$ (prime).

3. Test $n=5$, giving $2^5-1=31$ (prime).

4. Test the next prime, $n=7$, giving $2^7-1=127$ (prime).

5. Test the next prime, $n=11$, giving $2^{11}-1=2047$.

6. Show that $2047$ is composite by testing divisibility by small primes, revealing $2047=23\times 89$. Since $n=11$ yields a composite number, the statement is disproven.

1. Assume the opposite of the statement to be true: Assume that ${\log }_{2}5$ is a rational number.

2. Express the rational number as a fraction in its simplest form: ${\log }_{2}5=\frac{p}{q}$, where $p$ and $q$ are integers with no common factors, and $q\ne 0$.

3. Rewrite the logarithmic equation into its equivalent exponential form: $2^{\frac{p}{q}}=5$.

4. Raise both sides to the power of $q$ to eliminate the fraction from the exponent: $2^p=5^q$.

5. Analyze the parity (even/odd nature) of both sides. Since $p$ and $q$ must be non-zero positive integers, $2^p$ is always an even integer, and $5^q$ is always an odd integer.

6. State the contradiction: An even number cannot equal an odd number. Therefore, our initial assumption must be false, proving ${\log }_{2}5$ is irrational.

1. Establish the base case for $n=1$: $5^1-1=4$, which is a multiple of $4$. The statement therefore holds for $n=1$.

2. State the inductive assumption: Assume the statement is true for some integer $k$, such that $5^k-1=4M$ for some integer $M$.

3. Set up the inductive step to prove true for $n=k+1$: Consider the expression $5^{k+1}-1$.

4. Manipulate the expression algebraically to substitute the assumption: $5^{k+1}-1=5(5^k)-1=5(4M+1)-1$.

5. Expand and simplify the resulting expression: $20M+5-1=20M+4=4(5M+1)$.

6. Conclude: Since $5M+1$ is an integer, $5^{k+1}-1$ is a multiple of 4. As it holds for $n=1$ and $n=k⟹n=k+1$, it is true for all $n\in {\mathbb{Z}}^{+}$.