Unlocking the Code: A Guide to IB Math Proofs and Algebraic Secrets

By /

Hey everyone! Ever look at a math problem and feel like you’re being asked to prove something that’s just… obvious? Or maybe you see an algebraic identity and wonder, “How can two totally different-looking expressions always be the same?” Welcome to the world of mathematical proofs, one of the most powerful and, dare I say, beautiful parts of the IB Math curriculum. It’s not about crunching numbers; it’s about building a rock-solid, logical argument that leaves no room for doubt. Today, we’re going to crack two classic cases: one involving a strange property of even and odd numbers, and another that reveals a secret algebraic shortcut. Let’s get into it!

Table of Contents

The Divisibility Mystery: The Case of the Odd Couple

Our first mission is to tackle this statement: Prove that the sum of the squares of any even number and any odd number is never divisible by 2. In other words, we’re trying to prove that the result is always odd. It sounds plausible, but in math, we need more than a hunch. We need a formal proof.

Defining Our Suspects: Even & Odd Numbers

The first step in any proof like this is to translate our words into the language of algebra. We need universal definitions for even and odd numbers.

An even number is any integer that’s a multiple of 2. We can write this as $2n$, where $n$ is any integer ($n \in \mathbb{Z}$).

An odd number is always one more than an even number. We can write this as $2m + 1$, where $m$ is any integer ($m \in \mathbb{Z}$).

Pro Tip: Notice we use different variables, $n$ and $m$. This is crucial! It signifies that the even and odd numbers are independent and don’t have to be related to each other (like 4 and 9, not 4 and 5).

With our definitions in hand, let’s build the expression. We need the sum of their squares:

$ (even\ number)^2 + (odd\ number)^2 = (2n)^2 + (2m+1)^2 $

Next, we expand and simplify everything. Remember the rule for squaring a binomial, $(a+b)^2 = a^2 + 2ab + b^2$.

$ (2n)^2 + (2m+1)^2 = 4n^2 + (4m^2 + 4m + 1) $

$ = 4n^2 + 4m^2 + 4m + 1 $

Our goal is to check for divisibility by 2. This means we should try to factor a 2 out of the expression. Looking at the first three terms, they all have a common factor of 2.

$ = 2(2n^2 + 2m^2 + 2m) + 1 $

The ‘Let k’ Strategy

This next move is a classic in proofs. Look at the expression inside the parentheses: $(2n^2 + 2m^2 + 2m)$. Since $n$ and $m$ are integers, squaring them, multiplying by other integers, and adding them up will always result in another integer. To make our expression cleaner and reveal its true nature, we can substitute this entire chunk with a single variable, say $k$.

Let $k = 2n^2 + 2m^2 + 2m$, where $k \in \mathbb{Z}$.

Our expression magically simplifies to $2k + 1$.

And there it is! The expression $2k+1$ is the very definition of an odd number. Since our result is always odd, it can never be divisible by 2. Case closed. Q.E.D.

Your Turn: Proving the Product of an Even and Odd Number

Use the same logic to prove a different statement: Prove that the product of any even number and any odd number is always even.

Show Solution

1. Define terms: Let the even number be $2n$ and the odd number be $2m+1$.

2. Set up expression: We need their product: $(2n)(2m+1)$.

3. Expand and factor: $4nm + 2n = 2(2nm + n)$.

4. Conclude: Let $k = 2nm + n$. Since $n$ and $m$ are integers, $k$ is also an integer. The expression simplifies to $2k$, which is the definition of an even number. Therefore, the product is always even.

The Identity Crisis: Two Paths to the Same Truth

Next up, we’re asked to prove an algebraic identity: Show that $(x + 2)^2 – (x – 2)^2 \equiv 8x$. That triple-bar symbol $\equiv$ means this isn’t just an equation we solve for $x$; it’s an identity that holds true for any value of $x$ you can imagine. Our job is to show that the Left-Hand Side (LHS) can be algebraically manipulated to look exactly like the Right-Hand Side (RHS).

Method 1: The Brute-Force Expansion

The most straightforward way is to start with the more complicated side (the LHS) and expand everything. This is the ‘brute-force’ method. It always works, but you need to be careful with your algebra.

LHS = $(x + 2)^2 – (x – 2)^2$

First, expand each squared binomial:

$(x+2)^2 = x^2 + 4x + 4$

$(x-2)^2 = x^2 – 4x + 4$

Now, substitute them back in. And here comes the most common mistake in the book!

Beware the Negative Sign!

When you substitute, you are subtracting the entire second expression. You MUST use parentheses to keep it all together.

LHS = $(x^2 + 4x + 4) – (x^2 – 4x + 4)$

Forgetting these parentheses is a guaranteed way to get the wrong answer. Now, distribute the negative sign to every term in the second bracket.

LHS = $x^2 + 4x + 4 – x^2 + 4x – 4$

Now we just collect the like terms. The $x^2$ and $-x^2$ cancel out. The $+4$ and $-4$ also cancel. We’re left with:

LHS = $4x + 4x = 8x$

Since LHS = RHS, our identity is proven!

Method 2: The Elegant Difference of Squares Shortcut

The first method worked, but it was a bit messy. A skilled mathematician always looks for patterns and structure. Notice the form of the LHS: it’s something squared minus something else squared. This is a classic Difference of Squares.

Rule: Difference of Squares

This is straight from your IB Formula Booklet: $a^2 – b^2 = (a – b)(a + b)$. Recognizing this pattern can save you a ton of time and reduce the chance of algebraic errors.

In our problem, let $a = (x+2)$ and $b = (x-2)$. Now we just plug these into the formula:

LHS = $[(x+2) – (x-2)][(x+2) + (x-2)]$

Now simplify what’s inside each of the big brackets. Be careful with the signs in the first one!

First bracket: $[x+2-x+2] = [4]$

Second bracket: $[x+2+x-2] = [2x]$

Putting it together:

LHS = $[4][2x] = 8x$

Look at that! We got the same result in half the steps. Both methods are valid, but recognizing the Difference of Squares is a mark of an efficient problem-solver.

Your Turn: A New Identity to Conquer

Try this one on your own using both methods: Show that $(3x + 4)^2 – (3x – 4)^2 \equiv 48x$. Which method do you prefer?

Reveal Solution

Method 1 (Expansion):

LHS = $(9x^2 + 24x + 16) – (9x^2 – 24x + 16)$

LHS = $9x^2 + 24x + 16 – 9x^2 + 24x – 16$

LHS = $48x$. LHS = RHS.

Method 2 (Difference of Squares):

Let $a = (3x+4)$ and $b = (3x-4)$.

LHS = $[(3x+4) – (3x-4)][(3x+4) + (3x-4)]$

LHS = $[3x+4-3x+4][3x+4+3x-4]$

LHS = $[8][6x] = 48x$. LHS = RHS.

Key Takeaways & Final Thoughts

So, what have we learned? First, that proofs are all about building a logical story. You start with clear definitions, set up your expression, and manipulate it one step at a time until you arrive at your conclusion. Second, we saw that while brute force can work, keeping an eye out for elegant patterns like the Difference of Squares can make your life a whole lot easier. This isn’t just about passing exams; it’s about training your brain to see structure and find the most efficient path to a solution—a skill that’s valuable far beyond the math classroom.

Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.

Additional Practice

Question 1: Prove that the sum of two odd numbers is always an even number.

Show Answer

Solution: Let the two odd numbers be $2n+1$ and $2m+1$. Their sum is $(2n+1) + (2m+1) = 2n + 2m + 2 = 2(n+m+1)$. Let $k = n+m+1$. The sum is $2k$, which is the definition of an even number.

Question 2: Prove that the product of any two odd numbers is always an odd number.

Show Answer

Solution: Let the odd numbers be $2n+1$ and $2m+1$. Their product is $(2n+1)(2m+1) = 4nm + 2n + 2m + 1 = 2(2nm + n + m) + 1$. Let $k = 2nm + n + m$. The product is $2k+1$, which is the definition of an odd number.

Question 3: Show that $(x+5)^2 – (x-5)^2 \equiv 20x$.

Show Answer

Solution (Difference of Squares): $[(x+5)-(x-5)][(x+5)+(x-5)] = [10][2x] = 20x$.

Question 4: Prove that the sum of any three consecutive integers is divisible by 3.

Show Answer

Solution: Let the three consecutive integers be $n$, $n+1$, and $n+2$. Their sum is $n + (n+1) + (n+2) = 3n + 3 = 3(n+1)$. Since $n+1$ is an integer, the sum is a multiple of 3.

Question 5: Show that $(a+b)^2 + (a-b)^2 \equiv 2(a^2 + b^2)$.

Show Answer

Solution: LHS = $(a^2+2ab+b^2) + (a^2-2ab+b^2) = a^2+a^2+b^2+b^2+2ab-2ab = 2a^2 + 2b^2 = 2(a^2+b^2)$. LHS = RHS.

Question 6: Prove that $n^3 – n$ is divisible by 3 for any integer $n \geq 1$. (Hint: Factor the expression first).

Show Answer

Solution: Factor the expression: $n^3 – n = n(n^2 – 1) = n(n-1)(n+1)$. This can be rewritten as $(n-1)n(n+1)$, which is the product of three consecutive integers. In any set of three consecutive integers, one of them must be a multiple of 3. Therefore, their product must be divisible by 3.

Question 7: Prove that the square of any odd number is one more than a multiple of 8. (This is a bit trickier!)

Show Answer

Solution: Let the odd number be $2n+1$. Its square is $(2n+1)^2 = 4n^2 + 4n + 1 = 4(n^2+n)+1$. Now consider the term $n^2+n = n(n+1)$. This is the product of two consecutive integers, which we know must be even. So, we can write $n(n+1) = 2k$ for some integer $k$. Substituting this back in: $4(2k) + 1 = 8k + 1$. This is one more than a multiple of 8.

Question 8: Show that $(2x+y)^2 – (2x-y)^2 \equiv 8xy$.

Show Answer

Solution (Difference of Squares): Let $a=2x+y$ and $b=2x-y$. $[(2x+y)-(2x-y)][(2x+y)+(2x-y)] = [2y][4x] = 8xy$.

Question 9: Prove that the sum of the squares of two consecutive even numbers is a multiple of 4.

Show Answer

Solution: Let the two consecutive even numbers be $2n$ and $2n+2$. The sum of their squares is $(2n)^2 + (2n+2)^2 = 4n^2 + (4n^2 + 8n + 4) = 8n^2 + 8n + 4 = 4(2n^2 + 2n + 1)$. Since $2n^2+2n+1$ is an integer, the result is a multiple of 4.

IB Exam Style Question

Question 10: Prove that the difference between the squares of any two consecutive odd positive integers is always a multiple of 8.

Show Answer

Solution:

1. Define terms: Let the two consecutive odd positive integers be $2n+1$ and $2n+3$, where $n$ is a non-negative integer ($n \in \mathbb{Z}, n \ge 0$).

2. Set up expression: We need to find the difference of their squares. It’s conventional to subtract the smaller from the larger: $(2n+3)^2 – (2n+1)^2$.

3. Use Difference of Squares: Let $a = (2n+3)$ and $b = (2n+1)$.

The expression becomes $[(2n+3) – (2n+1)][(2n+3) + (2n+1)]$.

4. Simplify the brackets:

First bracket: $[2n+3-2n-1] = [2]$.

Second bracket: $[2n+3+2n+1] = [4n+4]$.

5. Combine and factor: The expression is now $[2][4n+4]$. We can factor a 4 out of the second bracket: $2 \cdot 4(n+1) = 8(n+1)$.

6. Conclude: Since $n$ is an integer, $(n+1)$ is also an integer. Therefore, $8(n+1)$ is, by definition, a multiple of 8. The statement is proven.

Shopping Cart
Scroll to Top