IB Maths AA SL Proof Questions

1. Expand the first binomial squared on the left-hand side (LHS): $(a-b{)}^2=a^2-2ab+b^2$.

2. Expand the second binomial squared: $(a+b{)}^2=a^2+2ab+b^2$.

3. Substitute these back into the LHS expression, using brackets to protect signs: $(a^2-2ab+b^2)-(a^2+2ab+b^2)$.

4. Distribute the negative sign through the second bracket: $a^2-2ab+b^2-a^2-2ab-b^2$.

5. Cancel the $a^2$ and $b^2$ terms.

6. Simplify to yield the final expression: $-2ab-2ab=$ $-4ab$, which equals the RHS.

1. Define three consecutive integers algebraically. Let the integers be $n$, $n+1$, and $n+2$, where $n\in \mathbb{Z}$.

2. Set up the sum of these integers: $\text{Sum}=n+(n+1)+(n+2)$.

3. Collect like terms together: $\text{Sum}=3n+3$.

4. Factorise the expression by factoring out the common divisor: $\text{Sum}=3(n+1)$.

5. State the logical conclusion: Since $n$ is an integer, $(n+1)$ is also an integer.

6. Conclude the proof: Because the sum is $3$ times an integer, it is always a multiple of 3.

1. Define two distinct odd integers. Let them be $2k+1$ and $2m+1$, where $k,m\in \mathbb{Z}$.

2. Set up the product of these two integers: $(2k+1)(2m+1)$.

3. Expand the brackets using FOIL: $4km+2k+2m+1$.

4. Factorise a $2$ out of the first three terms to isolate the even component: $2(2km+k+m)+1$.

5. State that since $k$ and $m$ are integers, the expression $(2km+k+m)$ must also evaluate to an integer. Let this be integer $P$.

6. Conclude the proof: The product can be written in the form $\mathbf{2}\mathbf{P}\mathbf{+}\mathbf{1}$, which is the mathematical definition of an odd number.