\( \DeclareMathOperator{cosec}{cosec} \)

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International Baccalaureate Mathematics

Number and Algebra

Syllabus Content 1.6

Simple deductive proof, numerical and algebraic; how to lay out a left-hand side to right-hand side (LHS to RHS) proof. The symbols and notation for equality and identity

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Here is an Advanced Starter on this statement:

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Furthermore

This video on Deduction is from Revision Village and is aimed at students taking the IB AA Maths Standard level course


Official Guidance, clarification and syllabus links:

Example: Show that \( \frac{1}{4} + \frac{1}{12} = \frac{1}{3} \). Show that the algebraic generalisation of this is \( \frac{1}{m+1} + \frac{1}{m^2+m} \equiv \frac{1}{m} \)

LHS to RHS proofs require students to begin with the left-hand side expression and transform this using known algebraic steps into the expression on the right-hand side (or vice versa).

Example: Show that \( (x-3)^2+5 \equiv x^2-6x+14 \).

Students will be expected to show how they can check a result including a check of their own results.


Example, Show that the algebraic generalisation of the following is true.

$$ \frac{1}{m+1} + \frac{1}{m^2+m} \equiv \frac{1}{m} $$

Let's start by finding a common denominator for the left-hand side of the equation:

$$ \frac{1}{m+1} + \frac{1}{m(m+1)} $$ $$ \frac{m + 1}{m(m+1)} $$

Now, let's simplify the expression:

$$ \frac{1}{m} $$

It is now evident that the original equation:

$$ \frac{1}{m+1} + \frac{1}{m^2+m} \equiv \frac{1}{m} $$

Is generally true, as the left-hand side simplifies to \( \frac{1}{m} \).

Thus, the algebraic generalisation provided does hold true for all values of \( m \).


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