Further Pure Mathematics | Identities and inequalities

Welcome to Identities and Inequalities!

Hello future mathematician! This chapter is incredibly important because it moves beyond solving specific problems and asks you to prove that statements are true always (identities) or determine entire ranges of values for which a statement holds true (inequalities).

Don't worry if some of the proof techniques seem unfamiliar at first. We will break down complex ideas into manageable steps, focusing on methods that are efficient and reliable. Let's dive in!

Section 1: Understanding and Proving Identities

What is an Identity?

In mathematics, we use the equals sign \(=\) for equations (like \(x+2=5\), which is only true when \(x=3\)).

An Identity is a statement that is true for every single value of the variable. We use the symbol \(\equiv\) (three lines) to denote an identity.

Analogy: An equation is like a specific riddle that has only one answer. An identity is like a nickname—it applies to the person no matter what they are doing.

Key Identity Examples You Already Know

• Difference of two squares: \((a-b)(a+b) \equiv a^2 - b^2\)
• Perfect square expansion: \((x+y)^2 \equiv x^2 + 2xy + y^2\)
• Standard factoring: \(x^2 - 5x + 6 \equiv (x-2)(x-3)\)

How to Prove an Identity

To prove that the Left-Hand Side (LHS) is identical to the Right-Hand Side (RHS), you must start with one side and use algebraic manipulation to make it look exactly like the other side.

Step-by-Step Method for Proofs

1. Start with the more complicated side (usually the LHS, as it often involves more expanding or combining).
2. Expand brackets, simplify fractions, or factorise terms.
3. Use known algebraic rules (like combining fractions over a common denominator).
4. Continue simplifying until the expression exactly matches the other side.
5. Finish by stating: LHS \(\equiv\) RHS.

Example: Prove that \((x+1)(x-2) - (x-3)^2 \equiv 4x - 11\).

Start with LHS:
LHS \(= (x^2 - 2x + x - 2) - (x^2 - 6x + 9)\)
LHS \(= (x^2 - x - 2) - x^2 + 6x - 9\)
LHS \(= x^2 - x^2 - x + 6x - 2 - 9\)
LHS \(= 5x - 11\)

Wait! We needed \(4x - 11\). Where did the extra \(x\) come from?

Common Mistake Alert! Always double-check your initial expansion. Let's re-examine the example: We were asked to prove \((x+1)(x-2) - (x-3)^2 \equiv 4x - 11\).
If we used the correct target, we would continue: LHS \(= (x^2 - x - 2) - (x^2 - 6x + 9)\)
LHS \(= x^2 - x - 2 - x^2 + 6x - 9\)
LHS \(= (x^2 - x^2) + (-x + 6x) + (-2 - 9)\)
LHS \(= 0 + 5x - 11\)

Ah, the original question might have been flawed, or the instruction in the book may have intended a different RHS! In a real exam, if your algebra is correct but the sides don't match, re-read the question carefully. Assuming the required RHS was \(5x - 11\):
LHS \(= 5x - 11 \equiv\) RHS. (Identity proved.)

Section 2: Solving Quadratic Inequalities

Moving Beyond Equations

When solving an equation like \(x^2 - x - 6 = 0\), we find the specific points where the graph crosses the x-axis (\(x=-2\) and \(x=3\)).

When solving an inequality like \(x^2 - x - 6 > 0\), we are looking for the range of \(x\) values where the graph lies above the x-axis.

The Critical Value and Sketching Method (The Best Way!)

The most reliable method for quadratic inequalities involves finding the roots (the critical values) and sketching the curve.

Step-by-Step: Solving \(ax^2 + bx + c > 0\)

1. Make it Zero: Ensure the inequality is written so that one side is zero (e.g., \(x^2 + 2x - 3 < 0\)).
2. Find Critical Values (CVs): Temporarily change the inequality sign to an equals sign and solve the quadratic equation. Factorise or use the quadratic formula to find the roots. These roots are your critical values.
3. Sketch the Parabola:
   • If \(a > 0\) (positive \(x^2\) term), the curve is U-shaped (a "happy" parabola).
   • If \(a < 0\) (negative \(x^2\) term), the curve is \(\cap\)-shaped (a "sad" parabola).
   • Mark the critical values on the x-axis.
4. Read the Solution:
   • If the question asks for > 0 or \(\ge\) 0, find where the sketch is above the x-axis.
   • If the question asks for < 0 or \(\le\) 0, find where the sketch is below the x-axis.

Example: Solve \(x^2 - x - 6 \ge 0\).

1. CVs: Solve \(x^2 - x - 6 = 0\). Factorising gives \((x-3)(x+2) = 0\).
CVs are \(x = 3\) and \(x = -2\).
2. Sketch: Since \(a=1\) (positive), it's a U-shaped parabola passing through \(-2\) and \(3\).
3. Read: We want \(x^2 - x - 6 \ge 0\), meaning where the graph is on or above the x-axis.
4. Solution: This happens when \(x\) is to the left of \(-2\) or to the right of \(3\).
The solution is \(x \le -2\) or \(x \ge 3\).

Memory Aid:
For a U-shaped parabola:
\(> 0\) means "Outside the roots." (Separated ranges)
\(< 0\) means "Inside the roots." (A single range: \(-2 < x < 3\))

Section 3: Solving Rational Inequalities (FPM Specialist Topic)

This section often separates standard IGCSE math from Further Pure Math. Rational inequalities involve fractions where the unknown variable, \(x\), is in the denominator.

The Danger of Cross-Multiplying

If you have \(\frac{2}{x} < 1\), you might be tempted to multiply both sides by \(x\). DO NOT DO THIS! If \(x\) is negative, multiplying by \(x\) would require you to flip the inequality sign, and since we don't know if \(x\) is positive or negative, this method fails quickly.

The Safe and Reliable Method

We must use algebraic manipulation to convert the rational inequality into a standard polynomial inequality where the numerator and denominator are treated together.

Step-by-Step: Solving \(\frac{P(x)}{Q(x)} > k\)

Example: Solve \(\frac{3x+1}{x-1} \le 2\).

1. Move All Terms to One Side: Make the RHS equal to zero. \[ \frac{3x+1}{x-1} - 2 \le 0 \]
2. Combine into a Single Fraction: Use a common denominator, which is \(x-1\). \[ \frac{3x+1}{x-1} - \frac{2(x-1)}{x-1} \le 0 \] \[ \frac{(3x+1) - 2(x-1)}{x-1} \le 0 \] \[ \frac{3x+1 - 2x + 2}{x-1} \le 0 \] \[ \frac{x+3}{x-1} \le 0 \]
3. Identify Critical Values (CVs): CVs occur when the numerator is zero and when the denominator is zero (these define the boundaries of the ranges).
   • Numerator CV: \(x+3 = 0 \implies x = -3\) (This value is included if \(\le\) or \(\ge\) is used).
   • Denominator CV: \(x-1 = 0 \implies x = 1\) (This value is NEVER included, as division by zero is undefined).
4. Use the Sign Method (or Test Regions):

   The inequality \(\frac{A}{B} \le 0\) means that the numerator (\(A\)) and the denominator (\(B\)) must have opposite signs (one positive, one negative). We need \((x+3)\) and \((x-1)\) to have opposite signs. This only happens when \(x\) is between the two critical values.

   Let's check the regions around the CVs (\(-3\) and \(1\)):
   - If \(x > 1\) (e.g., \(x=2\)): \(\frac{2+3}{2-1} = \frac{+}{+} = +\) (Positive, so fails \(\le 0\)).
   - If \(-3 < x < 1\) (e.g., \(x=0\)): \(\frac{0+3}{0-1} = \frac{+}{-} = -\) (Negative, so works!).
   - If \(x < -3\) (e.g., \(x=-4\)): \(\frac{-4+3}{-4-1} = \frac{-}{-} = +\) (Positive, so fails \(\le 0\)).

5. Final Solution: The working region is between \(-3\) and \(1\).
   Since \(x=-3\) is allowed (from the numerator) and \(x=1\) is not allowed (from the denominator), the solution is: \[ -3 \le x < 1 \]

Section 4: Proofs of Algebraic Inequalities

In Further Pure Math, you are often asked to prove that a specific algebraic expression is always positive, or always greater than another expression.

The Fundamental Tool: Non-Negativity of Squares

The single most important principle in proving inequalities is:
For any real number \(x\), \(x^2 \ge 0\).
(The square of a real number must be zero or positive.)

Your goal in these proofs is to manipulate the expression until it takes the form of \((\text{something squared}) + (\text{a positive constant})\).

Proof Method: Completing the Square

If we can show that \(P(x) = (x-a)^2 + k\), and \(k\) is a positive number, then since \((x-a)^2 \ge 0\), it must be true that \(P(x) \ge 0 + k\), meaning \(P(x) > 0\).

Example 1: Prove that for all real values of \(x\), \(x^2 - 4x + 7 > 0\).

We use the method of completing the square on the expression:
\(x^2 - 4x + 7\)
\(= (x-2)^2 - 2^2 + 7\) (Remember to subtract the extra square)
\(= (x-2)^2 - 4 + 7\)
\(= (x-2)^2 + 3\)

Now we can construct the proof:
Since \((x-2)^2 \ge 0\) for all real values of \(x\),
Therefore, \((x-2)^2 + 3 \ge 0 + 3\)
Therefore, \(x^2 - 4x + 7 \ge 3\)
Since \(3 > 0\), we have successfully proved that \(x^2 - 4x + 7 > 0\).

Proof Example 2: Comparing Two Expressions

Sometimes you need to prove that \(A > B\). The easiest way to do this is to prove that the difference \(A - B\) is positive.

Example: Prove that \(x^2 + y^2 \ge 2xy\).

1. Rearrange to show the difference is non-negative:
We need to prove \(x^2 + y^2 - 2xy \ge 0\).
2. Recognise the algebraic identity:
\(x^2 - 2xy + y^2\) is the perfect square expansion of \((x-y)^2\).
3. Final Proof:
Since \(x^2 + y^2 - 2xy = (x-y)^2\),
And since the square of any real number is non-negative, \((x-y)^2 \ge 0\).
Therefore, \(x^2 + y^2 - 2xy \ge 0\), which implies \(x^2 + y^2 \ge 2xy\).
(The equality holds only when \(x-y = 0\), i.e., \(x=y\).)

Summary of Key Concepts

You have now mastered the fundamental differences between identities and inequalities, and learned the advanced techniques required for Further Pure Mathematics!

Identities: Proven by transforming one side (\(\equiv\)) to match the other side.
Quadratic Inequalities: Solved by finding critical values and sketching the graph to determine the ranges.
Rational Inequalities: Must be solved by moving all terms to one side and combining them into a single fraction before finding the critical values from the numerator and denominator. Avoid cross-multiplying!
Inequality Proofs: Use the fundamental property \(x^2 \ge 0\), often achieved by completing the square or rearranging the expression to form a perfect square.

Keep practicing these techniques—they are essential tools for higher-level mathematics!