← | Table of contents | → |

Most students have been exposed to proofs of equations based on
the idea of *equality-preserving transformations*.

An equality-preserving transformation converts an expression into an equal one, typically by replacing a subexpression with an equal subexpression.

The simplest equality-preserving transformations involve arithmetic. For example, you know that (2)(14) = 28. Transforming expression (2)(14) into 28 is an equality-preserving transformation.

You also know that *x*(*y*+*z*) = *x**y* + *x**z*.
That means

(5w) + (x(y+z)) = (5w) + (xy+xz)

since you can perform an equality-preserving transformation on a subexpression.

One way to prove that an equation is true is to start with one side (say, the left-hand side) and to convert it, by a sequence of equality-preserving transformations, into the other side. But remember that a proof must be easy to check, so each step deserves a justification. We keep the steps elementary, based on known facts of algebra. Also, a proof should always start with a statement of the claim that is being proved. Here is an example.

Claim.For everyxandy, (x+y)(2x+y) = 2x^{2}+ 3xy+y^{2}Proof.( x+y)(2x+y)= x(2x+y) +y(2x+y)(by the distributive law, ( a+b)c=ac+bc)= ( x(2x) +x(y)) +y(2x+y)(by the distributive law, a(b+c) =ab+bc)= ( x(2x) +x(y)) + (y(2x) +y(y))(by the distributive law, a(b+c) =ab+bc)= x(2x) +x(y) +y(2x) +y(y)(Since addition is associative (( a+b) +c=a+ (b+c)),

the parentheses are irrelevant)= 2 x^{2}+xy+ 2xy+y^{2}(by associativity and commutivity of multiplication) = 2 x^{2}+ 3xy+y^{2}(since a+ 2a= 3a)Having converted (

x+y)(2x+y) to 2x^{2}+ 3xy+y^{2}by a sequence of equality-preserving transformations, we conclude that the claim must be true. Also, notice that the transformations do not depend on the values ofxandy, so they must work for every possible choice of values ofxandy.♦ (end of proof)

Just as a matter of presentation, we can save some horizontal space by moving the right-hand sides over, as follows.

Claim.For everyxandy, (x+y)(2x+y) = 2x^{2}+ 3xy+y^{2}Proof.( x+y)(2x+y)= x(2x+y) +y(2x+y)(by the distributive law, ( a+b)c=ac+bc)= ( x(2x) +x(y)) +y(2x+y)(by the distributive law, a(b+c) =ab+bc)= ( x(2x) +x(y)) + (y(2x) +y(y))(by the distributive law, a(b+c) =ab+bc)= x(2x) +x(y) +y(2x) +y(y)(Since addition is associative (( a+b) +c=a+ (b+c)),

the parentheses are irrelevant)= 2 x^{2}+xy+ 2xy+y^{2}(by associativity and commutivity of multiplication) = 2 x^{2}+ 3xy+y^{2}(since a+ 2a= 3a)♦

This proof is much more detailed than you are accustomed to. When you are working
in an area, such as elementary algebra, that is familiar to both you and to the reader,
you can often shorten it. In fact, we will justify some steps by saying
*by algebra* when they should be clear to anyone with an understanding of
common algebraic rules. But be careful when working in an unfamiliar area. Trying to
shorten a proof by doing more than one step at a time is one of the most common ways
of introducing mistakes.

What if the goal had been to show that
2*x*^{2} + 3*x**y* + *y*^{2} =
(*x*+*y*)(2*x*+*y*),
for every *x* and *y*. You might start with the left-hand side
and try to convert it into the right-hand side by equality-preserving
transformations. But doing that requires more inspiration about what
to do than the other direction. For example, you need to realize that
a good place to start is to replace 3*x**y* by
*x**y* + 2*x**y*, which is certainly not obvious.

Sometimes, one conversion direction is easier than the other. But sometimes
neither direction is obvious. An idea then is to convert each side of
the equation into some standard form and to compare the results.
For example, suppose that the claim is that
(*x*+*y*)(2*x*+*y*) =
3(*x*^{2} + *x**y*) − (*x*^{2} − *y*^{2}), for every *x* and *y*.

Claim.For everyxandy, (x+y)(2x+y) = 3(x^{2}+xy) − (x^{2}−y^{2}).Proof.Subclaim 1.For everyxandy, (x+y)(2x+y) = 2x^{2}+ 3xy+y^{2}.Proof of subclaim 1.See proof above.

Subclaim 2.For everyxandy, 3(x^{2}+xy) − (x^{2}−y^{2}) = 2x^{2}+ 3xy+y^{2}.Proof of subclaim 2.3( x^{2}+xy) − (x^{2}−y^{2})= 3 x^{2}+ 3xy−x^{2}+y^{2}(by algebra) = 2 x^{2}+ 3xy+y^{2}(by algebra) ♦