6.01a - Identities and algebraic arguments

6.01a - Identities and algebraic arguments

Two expressions can look different yet give the same result for every input. Learn to distinguish an identity from an equation to solve, show equivalence using algebra and build arguments that explain why a result holds.

Equations and identities

A number trick says: choose a number, multiply it by 3, add 12, divide by 3, then subtract your starting number. Why does the answer seem to be 4 every time? Trying more starting numbers gives evidence; algebra will explain the result for all of them.

First compare two equalities. 2(x+3)=2x+62(x+3)=2x+6 is true at x=1x=1, at x=−4x=-4 and at every other real input, because expanding the bracket gives the right side. By contrast, 2x+3=112x+3=11 is true only when x=4x=4.

Identity

An equation states that two expressions are equal. An identity is an equality true for every permitted value of its variables. We can use ≡\equiv to emphasise an identity: 2(x+3)≡2x+62(x+3)\equiv2x+6. Ordinary == is also used when showing the steps of an identity.

In questions that ask you to choose between “equation” and “identity”, use “identity” for an equality true for all permitted values, and “equation” for an equality that holds only for particular values. Some equations have no solution: x+1=x+2x+1=x+2 would require 1=21=2, so it is never true and is not an identity.

The word “permitted” matters. For example, xx=1\frac{x}{x}=1 is true for every nonzero xx, but the left side is undefined at zero. Simplifying must not silently allow a value that the original expression excluded.

Key idea

Solving an equation asks which inputs make it true. Establishing an identity explains why it is true for every permitted input.

Choose the statement whose equality survives every real input.

Showing expressions are equivalent

To show two expressions are equivalent, transform one into the other by valid algebraic steps. This works for an unspecified input, so it establishes more than a table of successful numerical checks.

Equivalent expressions

Equivalent expressions have the same value for every input for which they are being compared. Expanding, collecting like terms and factorising can reveal this equivalence, while preserving any original restrictions.

When one side has brackets and the other is expanded, starting from the bracketed side usually gives the clearest argument. Keep the target in sight without assuming it is already true.

Worked example 1: showing a quadratic identity

Question. Show that (x+3)2+1=x2+6x+10(x+3)^2+1=x^2+6x+10 for every real xx.

Plan. Expand the square on the left, then collect like terms.

(x+3)2+1=(x+3)(x+3)+1write the square as a product=x2+3x+3x+9+1multiply each term=x2+6x+10collect like terms\begin{aligned} (x+3)^2+1&=(x+3)(x+3)+1 &&\text{write the square as a product} \\ &=x^2+3x+3x+9+1 &&\text{multiply each term} \\ &=x^2+6x+10 &&\text{collect like terms} \end{aligned}

Answer. The expressions are equivalent for every real xx, because the left side simplifies to the right side without choosing a value of xx.

Check. At x=2x=2, the two original forms give 52+1=265^2+1=26 and 4+12+10=264+12+10=26. This checks the arithmetic in one case; the expansion supplies the general argument.

Change only the sign inside the bracket. The two cross-products change sign, while the square of three stays positive.

Worked example 2: changing the middle term

Question. Show that (x−3)2+1=x2−6x+10(x-3)^2+1=x^2-6x+10 for every real xx.

Plan. Expand the two factors, keeping both negative cross-products.

(x−3)2+1=(x−3)(x−3)+1write the product=x2−3x−3x+9+1expand=x2−6x+10collect like terms\begin{aligned} (x-3)^2+1&=(x-3)(x-3)+1 &&\text{write the product} \\ &=x^2-3x-3x+9+1 &&\text{expand} \\ &=x^2-6x+10 &&\text{collect like terms} \end{aligned}

Answer. The left side equals the right side for every real xx.

Check. At x=2x=2, the first form gives (−1)2+1=2(-1)^2+1=2, and the second gives 4−12+10=24-12+10=2. The positive constant agrees with the product (−3)(−3)(-3)(-3).

You can also work towards a common expression from both sides. State what each side simplifies to, rather than writing a chain containing an unproved equality. When a “show that” question supplies the final expression, the useful evidence is the algebra connecting the start to that expression.

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