Identities and Algebraic Proof
1. Equations vs. Identities
Equation: Only true for specific values of $x$. You solve it to find out what $x$ is.
Example: $2x + 1 = 7$ (This is only true when $x = 3$).
Identity: True for all values of $x$. It just means two expressions are written differently but mean the exact same thing. We use the $\equiv$ symbol to show this.
Example: $2(x + 3) \equiv 2x + 6$ (This is true whether $x$ is 1, 10, or a million).
2. Proving Equivalent Expressions
When an exam asks you to "Show that..." two expressions are equivalent, you must pick one side and manipulate it until it looks exactly like the other side.
The Golden Rule of Proofs:
DO NOT move things across the equals sign like you are solving an equation! Treat the left side and the right side as two completely separate islands. Work on the "messiest" side until it matches the neat side.
Example: Show that $(x+4)^2 - 16 \equiv x(x+8)$
Step 1: Start with the Left Hand Side (LHS)
LHS = $(x+4)(x+4) - 16$
Step 2: Expand the brackets
LHS = $x^2 + 4x + 4x + 16 - 16$
Step 3: Simplify
LHS = $x^2 + 8x$
Step 4: Factorise to match the Right Hand Side (RHS)
LHS = $x(x+8)$
Since LHS = RHS, the identity is proven.
3. Constructing Algebraic Proofs
To prove something is always true for numbers, we can't just test a few numbers (like 2, 3, and 4) because we haven't proven it works for 1,000,000. We must use algebra!
The Core Vocabulary of Proof:
Let $n$ be any integer (whole number).
- An Even Number: $2n$ (because multiplying by 2 always makes it even).
- An Odd Number: $2n + 1$ (an even number plus 1 is always odd).
- Consecutive Integers: $n$, $n+1$, $n+2$, $n+3$...
- Consecutive Even Integers: $2n$, $2n+2$, $2n+4$...
- A Multiple of 3: $3n$ (or $3(\text{something})$).
Example: Prove that the sum of two consecutive integers is always odd.
Step 1: Define the numbers
Let the first integer be $n$.
Let the second consecutive integer be $n+1$.
Step 2: Do the math (find the sum)
Sum = $n + (n+1)$
Sum = $2n + 1$
Step 3: Write a concluding sentence
Since $2n$ is always even, $2n+1$ must always be odd. Therefore, the sum is always odd.
Practice Questions & Solutions
Try these on paper first. Remember: write clear steps!
Q1: Is $3x + 4x \equiv 7x$ an equation or an identity?
It is true for every value of $x$ you could possibly choose.
Answer: It is an identity.
Q2: Show that $5(x+2) - 3(x+1) \equiv 2x + 7$
Start with the LHS and expand the brackets:
$= 5x + 10 - 3x - 3$
Collect like terms ($5x - 3x$ and $10 - 3$):
$= 2x + 7$
This matches the RHS, so it is proven.
Q3: Prove that the sum of three consecutive integers is always a multiple of 3.
Let the integers be $n$, $n+1$, and $n+2$.
Sum $= n + (n+1) + (n+2)$
Sum $= 3n + 3$
Factorise out the 3 to prove it's a multiple of 3:
$= 3(n + 1)$
Since the expression is a multiple of 3, the proof is complete.
Q4: Prove that the sum of two odd numbers is always even.
Let the first odd number be $2n+1$.
Let the second odd number be $2m+1$. (We use $m$ because it might be a different odd number).
Sum $= (2n+1) + (2m+1)$
Sum $= 2n + 2m + 2$
Factorise out a 2:
$= 2(n + m + 1)$
Because it is a multiple of 2, the result is always even.
Q5: Show that $(x+3)^2 \equiv x^2 + 6x + 9$
Expand the LHS using double brackets:
$(x+3)(x+3)$
$= x^2 + 3x + 3x + 9$
$= x^2 + 6x + 9$
This exactly matches the RHS.
Q6: Prove algebraically that the square of an even number is always a multiple of 4.
Let the even number be $2n$.
Square it: $(2n)^2$
$= (2n) \times (2n)$
$= 4n^2$
Since $4n^2$ is a multiple of 4 ($4 \times n^2$), the proof is complete.
Q7: Show that $(n+2)^2 - n^2 \equiv 4n + 4$
Expand the LHS:
$(n+2)(n+2) - n^2$
$= (n^2 + 2n + 2n + 4) - n^2$
$= n^2 + 4n + 4 - n^2$
The $n^2$ terms cancel out:
$= 4n + 4$
This matches the RHS.
Q8: Prove that the product of two even numbers is always a multiple of 4.
Let the two even numbers be $2n$ and $2m$.
Product (multiply them): $(2n) \times (2m)$
$= 4nm$
Since there is a factor of 4, the result is always a multiple of 4.
Q9: Show that $\frac{6x+12}{3} \equiv 2(x+2)$
Start with the LHS. Divide both terms on the top by 3:
$\frac{6x}{3} + \frac{12}{3}$
$= 2x + 4$
Now factorise this expression:
$= 2(x + 2)$
This matches the RHS.
Q10: Challenge! Prove that the difference between the squares of two consecutive integers is always an odd number.
Let the integers be $n$ and $n+1$.
Their squares are $n^2$ and $(n+1)^2$.
Difference (biggest minus smallest):
$(n+1)^2 - n^2$
Expand the bracket:
$(n^2 + 2n + 1) - n^2$
Simplify:
$= 2n + 1$
Since $2n$ is always even, $2n+1$ is always odd. Proof complete.