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Algebraic proof

Free Printable Algebraic proof PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 8+)

Download free algebraic proof worksheets (PDF), GCSE maths past paper questions and worked solutions for Algebraic proof (8C). This is a Grade 8+ algebra topic from the GCSE questions by topic collection for Foundation and Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from IGCSE MT, Maths Genie, Corbett Maths, Maths Made Easy, Dr Frost, Naiker Maths, 1st Class Maths, Metatutor, Maths Takeaway, Dr Austin, Maths4EveryOne, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our algebraic proof revision notes and worked examples with diagrams, try real GCSE past paper algebraic proof questions with step-by-step solutions, or download a PDF of all algebraic proof worksheet links.

Free Algebraic proof PDF Worksheets & Answers

Grade 8+ • Topic 8C
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GCSE Maths Grade 8+ Revision Notes & Method

Guide to Algebraic proof: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

Never use numerical examples to prove an algebraic statement—you must use general algebra for any integer n: an even number is 2n, an odd number is 2n + 1, consecutive integers are n, n + 1, n + 2, and consecutive odd numbers are 2n + 1, 2n + 3.

Expand and simplify the expression, then factorise out the required multiple (e.g. write 8n + 4 as 4(2n + 1) to prove it is a multiple of 4) and finish with a concluding sentence explaining why the result holds for all integers n.

Example 1: Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.

Solution:

Let the consecutive odd numbers be 2n + 1 and 2n + 3:
(2n + 3)2 − (2n + 1)2 = (4n2 + 12n + 9) − (4n2 + 4n + 1)
= 8n + 8 = 8(n + 1)
Since n + 1 is an integer, 8(n + 1) is always a multiple of 8.

Example 2: Prove that n2 − 6n + 13 is positive for all real values of n.

Solution:

Complete the square on n2 − 6n + 13:
= (n − 3)2 − 9 + 13 = (n − 3)2 + 4
Since any squared real number is ≥ 0, (n − 3)2 ≥ 0
Therefore (n − 3)2 + 4 ≥ 4 > 0, so it is always positive.
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Algebraic proof Questions & Answers

Past Paper Example 1 IGCSE Jun 2019 2HR Q17 4 marks
The table gives information about the first six terms of a sequence of numbers.
Term numberTerm of sequence11 × 2222 × 3233 × 4244 × 5255 × 6266 × 72
Prove algebraically that the sum of any two consecutive terms of this sequence is always a square number.
Worked Solution & Mark Scheme (4 marks)
nth term + (n+1)th term = ½n(n+1) + ½(n+1)(n+2) = ½(n² + n + n² + 3n + 2) = ½(2n² + 4n + 2) = n² + 2n + 1 = (n + 1)² M3 C1
Answer: n(n+1)2 + (n+1)(n+2)2 = (n+1)(2n+2)2 = (n+1)2, which is a square number
Past Paper Example 2 GCSE Nov 2019 3H Q15 3 marks
Prove algebraically that the sum of the squares of any two consecutive even numbers is always a multiple of 4
Worked Solution & Mark Scheme (3 marks)
Let the two consecutive even numbers be 2n and 2n + 2 M2
(2n)2 + (2n + 2)2 = 4n2 + 4n2 + 8n + 4 = 8n2 + 8n + 4 = 4(2n2 + 2n + 1), which is a multiple of 4. C1
Answer: (2n)2 + (2n + 2)2 = 8n2 + 8n + 4 = 4(2n2 + 2n + 1)
Past Paper Example 3 GCSE Nov 2017 1H Q17 2 marks
Work out
n is an integer.
Prove algebraically that the sum of 12n(n + 1) and 12(n + 1)(n + 2) is always a square number.
Worked Solution & Mark Scheme (2 marks)
Factorising 1/2(n + 1)[n + (n + 2)] = 1/2(n + 1)(2n + 2) = (n + 1)(n + 1) = (n + 1)², which is the square of an integer. M1 C1
Answer: 12(n2 + n) + 12(n2 + 3n + 2) = 12(2n2 + 4n + 2) = n2 + 2n + 1 = (n + 1)2, which is a square number

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