Proof by mathematical induction

A proof by mathematical induction is a powerful method that is used to prove that a conjecture (theory, proposition, speculation, belief, statement, formula, etc...) is true for all cases. 

Just because a conjecture is true for many examples does not mean it will be for all cases.

In order to show that the conjecture is true for all cases, we can prove it by mathematical induction as outlined below.

Important notes and explanations about a proof by mathematical induction

In 1., you are trying to show that the conjecture is true for specific values. You are free to do this test with just one value or fifty values of your choice or more.

However, showing it is true for one million values or more still does not prove it will be true for all values. This is a very important observation!

In 2., since you have already shown that the conjecture is true for one or more values, it is logical to suppose or assume it is true for n = k or generally speaking.

We usually use the assumption that we make here to complete or prove 3.

In 3., you finally show it is true for any values. Notice that 2. did not show it is true for any values.

An example showing how to do a proof by mathematical induction

Show that for all n, 2 + 4 + 6 + ... + 2n = n ( n + 1)

Step # 1:

Show the equation is true for n = 1, n = 2, .......

There is a pitfall to avoid here.

n = 1 means the first value of the expression on the left side. In this case 2

n = 2 means the first two values of the expression on the left side. In this case 2 + 4

n = 3 means the first three values of the expression on the left side. In this case 2 + 4 + 6

Thus, showing the equation 2 + 4 + 6 + ... + 2n = n ( n + 1) is true for n = 4 means that we have to show that 2 + 4 + 6 + 8 = 4 (4 + 1)

2 + 4 + 6 + 8 = 6 + 6 + 8 = 12 + 8 = 20 and 4 (4 + 1) = 4 × 5 = 20

Since the left side is equal to the right side (20 = 20) , step # 1 is done. It is not necessary to choose other values although you could do it just for fun and to prove to yourself that it will work for other values.

Step # 2:

Suppose the equation is true for n = k

Just replace n by k.

2 + 4 + 6 + ... + 2k = k ( k + 1)

Step # 3:

Prove the equation is true for n = k + 1

This is the toughest part of proof by mathematical induction. Things can get really tricky here. Not in this problem though!

At this point, you need to write down what it means for the equation to be true for n = k + 1

Be careful! Just because you wrote down what it means does not mean that you have proved it. This is another pitfall to avoid when working on a proof by mathematical induction.

Here is what it means:

After you replace k by k+1, you get :

2 + 4 + 6 + ... + 2 × (k + 1) = k+1 ( k + 1 + 1)

2 + 4 + 6 + ... + 2 × ( k + 1) = k+1 ( k + 2)

2 + 4 + 6 + ... + 2 × ( k + 1) = ( k + 1 ) × ( k + 2)

Let's give you a recap because you may have lost tract of what we are trying to do here.

We have not proved anything yet. The equation 2 + 4 + 6 + ... + 2 × ( k + 1) = ( k + 1 ) × ( k + 2) is just what it means for the equation to be true for n = k + 1

We are now ready to complete the proof by mathematical induction by using the hypothesis in step # 2.

starting with the hypothesis, 2 + 4 + 6 + ... + 2k = k ( k + 1)

Say to yourself, " What does the next term look like? "

Since the last term now is 2k, the next term should be 2 × ( k + 1)

Add 2 × ( k + 1) to both sides of the hypothesis

2 + 4 + 6 + ... + 2k + 2 × ( k + 1) = k ( k + 1) + 2 × ( k + 1)

                                                = k2 + k + 2k + 2

                                                 = k2 + 3k + 2

Since 2 = 1 × 2 and 1 + 2 = 3,

k2 + 3k + 2 = ( k + 1) × ( k + 2)

Therefore, 2 + 4 + 6 + ... + 2k + 2 × ( k + 1) = ( k + 1) × ( k + 2) and the proof by mathematical induction is complete!

The above is a well explained and solid proof by mathematical induction. Study it well!

Algebra proofs



What Do You Learn in 10th Grade Math?

In 10th grade math, students typically expand their knowledge in geometry with more complex theorems and proofs, and dive deeper into algebra with quadratic equations, functions, and inequalities. They are also introduced to trigonometry, focusing on the relationships between angles and sides of triangles.

Introduction

10th grade is a pivotal year in a student’s mathematical journey. Foundations from previous years are strengthened, and more advanced concepts are introduced.

This guide explores the key topics in 10th-grade math, with clear explanations and solved numerical examples.

Key Concepts and Skills in 10th Grade Math

Algebraic Mastery

Tenth graders deepen their algebra skills in equations, inequalities, and functions. Key topics include:

Quadratic Equations

Solving quadratic equations by factoring, completing the square, and using the quadratic formula. Example: Solve 2x² – 5x + 3 = 0.

Systems of Equations

Solving systems using substitution or elimination. Example:

2x + 3y = 10
4x – y = 5

Polynomial Operations

Addition, subtraction, multiplication, and division of polynomials.

Geometry and Trigonometry

Trigonometric Functions

Introduction to sine, cosine, and tangent. Students apply trigonometry to right triangles and periodic functions.

Figure-1: Generic Trigonometric Function

Geometry Proofs

Using deductive reasoning to prove statements about geometric figures.

Advanced Functions and Graphing

Exponential and Logarithmic Functions

Understanding growth/decay and solving exponential and logarithmic equations.

Graphing Functions

Graphing linear, quadratic, exponential, and trigonometric functions, including transformations.

Figure-2: Generic Function

Numerical Examples

Example 1: Quadratic Equations

Problem: Solve 2x² – 5x + 3 = 0.

Solution

Using the quadratic formula: x = (-b ± √(b² – 4ac)) / (2a)

a = 2, b = -5, c = 3

x = (5 ± √(25 – 24)) / 4 = (5 ± 1)/4

x₁ = 3/2, x₂ = 1

Example 2: Trigonometric Functions

Problem: Find sin(θ) where opposite = 4 and hypotenuse = 5.

Solution

sin(θ) = 4/5 = 0.8

Example 3: Systems of Equations

Problem: Solve the system:

2x + 3y = 10
4x – y = 5

Solution

Multiply the second equation by 3:

12x – 3y = 15

Add to the first equation:

14x = 25 → x = 25/14

Substitute to find y:

y = -15/7

Example 4: Exponential Equations

Problem: Solve 2(x-1) = 8.

Solution

8 = 2³, so:

x – 1 = 3 → x = 4

Example 5: Polynomial Operations

Problem: Simplify (3x² – 2x + 5) + (2x² + 4x – 1).

Solution

5x² + 2x + 4

Example 6: Geometry Proof

Problem: Prove that opposite angles of a parallelogram are congruent.

Solution

Consider parallelogram ABCD where AB ∥ CD and AD ∥ BC.

Using Alternate Interior Angles Theorem and properties of parallelograms, ∠A ≅ ∠C and ∠B ≅ ∠D.

Example 7: Exponential Growth

Problem: A bacteria colony doubles every hour. Starting with 100 bacteria, how many after 5 hours?

Solution

N(5) = 100 × 2⁵ = 100 × 32 = 3200 bacteria.

Example 8: Trigonometric Identity

Problem: Prove tan²(θ) + 1 = sec²(θ).

Solution

From Pythagorean identity: sin²(θ) + cos ²(θ) = 1 → 1 + tan²(θ) = sec ²(θ).

Conclusion

10th-grade mathematics builds critical thinking and problem-solving skills. Mastery of algebra, geometry, trigonometry, and functions prepares students for higher-level math and real-world applications.

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