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Problem Solving - Mathematical reasoning and proof

Grade 4IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Conjectures: Making a mathematical statement that you think is true but has not yet been proven.

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Counter-examples: Finding a single example that proves a general statement is false.

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Deductive Reasoning: Using known facts and logic to arrive at a certain conclusion.

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Proof by Exhaustion: Checking every single possible case to prove a statement is true for a specific set.

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Systematic Listing: Organizing data or possibilities in a logical order (e.g., a table) to ensure no outcomes are missed.

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Working Backwards: Starting from a known result to find the initial starting point.

📐Formulae

Even Number = 2n2n (where nn is an integer)

Odd Number = 2n+12n + 1 or 2n−12n - 1

Sum of interior angles in a triangle = 180∘180^{\circ}

General term of a sequence = a+(n−1)da + (n-1)d (for arithmetic progressions)

💡Examples

Problem 1:

Prove or disprove the statement: 'The sum of any two odd numbers is always an odd number.'

Solution:

Disproved. Let's test two odd numbers: 3+5=83 + 5 = 8. 8 is an even number.

Explanation:

To disprove a general statement, you only need to provide one counter-example. Since 3+5=83 + 5 = 8 and 8 is even, the statement 'The sum is always odd' is false.

Problem 2:

I am thinking of a number. If I multiply it by 3 and then add 7, the result is 22. What is the number?

Solution:

22−7=1522 - 7 = 15; 15÷3=515 \div 3 = 5. The number is 5.

Explanation:

This uses the 'Working Backwards' strategy. We reverse the operations in the opposite order: subtraction instead of addition, and division instead of multiplication.

Problem 3:

How many different ways can you make 10 cents using only 1-cent, 2-cent, and 5-cent coins?

Solution:

10 ways: (10x1), (8x1, 1x2), (6x1, 2x2), (4x1, 3x2), (2x1, 4x2), (5x2), (5x1, 1x5), (3x1, 1x2, 1x5), (1x1, 2x2, 1x5), (2x5).

Explanation:

This requires 'Systematic Listing'. We start with the largest coin (5-cent) and work down to ensure all combinations are found without skipping any.

Problem 4:

Prove that the sum of two even numbers is always even.

Solution:

Let the first even number be 2n2n and the second be 2m2m. 2n+2m=2(n+m)2n + 2m = 2(n + m). Since any number multiplied by 2 is even, the sum is even.

Explanation:

This is a simple algebraic proof. By factoring out a 2, we demonstrate that the result fits the definition of an even number (2×any integer2 \times \text{any integer}).