Example (a) Factorise \( 4x^3 + 18x^2 + 23x + 9 \). \( \begin{align} \displaystyle4x^3 + 18x^2 + 23x + 9 &= 4x^3 + 4x^2 + 14x^2 + 23x + 9 \\&= 4x^2 (x+1) + 14x^2 + 14x + 9x + 9 \\&= 4x^2 (x+1) + 14x(x+1) + 9(x+1) \\&= (x+1)(4x^2 + 14x + […]
Tag Archives: Proof
Mathematical Induction Regarding Factorials
Prove by mathematical induction that for all integers \( n \ge 1 \) ,$$ \dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \cdots + \dfrac{n}{(n+1)!} = 1-\dfrac{1}{(n+1)!}$$ Step 1 Show it is true for \( n=1 \).\( \begin{align} \displaystyle\text{LHS } &= \dfrac{1}{2!} = \dfrac{1}{2} \\\text{RHS } &= 1-\dfrac{1}{2!} \\&= 1-\dfrac{1}{2} \\&= \dfrac{1}{2} \end{align} \)Thus, the statement is […]
Proof by Contradiction
$\textbf{Introduction to Proof by Contradiction}$ The basic idea of $\textit{Proof by Contradiction}$ is to assume that the statement that we want to prove is $\textit{false}$, and then show this assumption leads to nonsense. We then conclude that it was wrong to assume the statement was $\textit{false}$, so the statement must be $\textit{true}$. As an example […]
Mathematical Induction Inequality
Mathematical Induction Inequality is being used for proving inequalities. It is quite often applied for subtraction and/or greatness, using the assumption in step 2. Let’s take a look at the following hand-picked examples. Basic Mathematical Induction Inequality Prove \( 4^{n-1} \gt n^2 \) for \( n \ge 3 \) by mathematical induction. Step 1: Show […]
Best Examples of Mathematical Induction Divisibility
Mathematical Induction Divisibility can be used to prove divisibility, such as divisible by 3, 5 etc. Same as Mathematical Induction Fundamentals, hypothesis/assumption is also made at step 2. Basic Mathematical Induction Divisibility Prove \( 6^n + 4 \) is divisible by \( 5 \) by mathematical induction, for \( n \ge 0 \). Step 1: […]
Mathematical Induction Fundamentals
The Mathematical Induction Fundamentals are defined for applying 3 steps: step 1 for showing its initial ignite, step 2 for making an assumption, and step 3 for showing it is true based on the assumption. Make sure the Mathematical Induction Fundamentals should be used only when the question asks to use it. Basic Mathematical Induction […]
Mathematical Induction Inequality Proof with Factorials
Worked Example Prove that \( (2n)! > 2^n (n!)^2 \) using mathematical induction for \(n \ge 2 \). Step 1: Show it is true for \( n =2 \).\( \begin{aligned} \require{AMSsymbols} \require{color}\text{LHS } &= (2 \times 2)! = 16 \\\text{RHS } &= 2^2 \times (2!) = 8 \\\text{LHS } &> { RHS} \\\end{aligned} \)\( \therefore […]
Inequality Proofs
Inequality Proof can be done in many ways. For proving \(A \ge B \), one of the easiest ways is to show \(A – B \ge 0 \). Worked Example of Inequality Proofs If \(a, b\) and \(c\) are positive real numbers and \(a+b\ge c \), prove that \( \displaystyle \frac{a}{1+a} + \frac{b}{1+b} \ge \frac{c}{1+c} […]
Absolute Value Inequalities
Absolute Value Inequalities are usually proved by the absolute value being greater than or equal to a certain value. The square of the value is equal to the square of its absolute value. Proof of Absolute Value Inequalities Prove \(|a| + |b| \ge |a+b|\). \( \begin{aligned} \require{AMSsymbols} \require{color}|a| &\ge a \text{ and } |b| \ge […]
Inequalities using Arithmetic Mean Geometric Mean
The Arithmetic Mean of \(a\) and \(b\) is always greater than or equal to the Geometric Mean of \(a\) and \(b\), for all positive real numbers with equality if and only if \(a = b\). This is also called AM-GM (Arithmetic Mean Geometric Mean) inequality.\(\require{color}\)$$ \begin{aligned}\frac{a + b}{2} \ge \sqrt{ab} \text{ or } a + […]
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