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# 1.4: Proving Identities

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There are many methods that one can use to prove an identity. The simplest is to use algebraic manipulation, as we have demonstrated in the previous examples. In an algebraic proof, there are three acceptable approaches:

• From left to right: expand or simplify the left-hand side until you obtain the right-hand side.

• From right to left: expand or simplify the right-hand side until you obtain the left-hand side.

• Meet in the middle: expand or simplify the left-hand side and the right-hand side separately until you obtain the same result from both sides.

Example $$\PageIndex{1}$$

To prove that $x^3-y^3 = (x-y) (x^2+xy+y^2),$ we start from the right-hand side, because it is more complicated than the left-hand side. The proof proceeds as follows:

Solution

\begin{aligned}(x-y)(x^2+xy+y^2) &=& x^3-x^2y+x^2y-xy^2+xy^2-y^3 \\ &=& x^3-y^3.\end{aligned}\label{eg:provingID-01} Remember: start from one side and work on it until you obtain the other side.

Example $$\PageIndex{2}$$

The following “proof” of $x^4+x^2y^2+y^4 = (x^2+xy+y^2) (x^2-xy+y^2)$ is incorrect:

$\begin{eqnarray*} x^4+x^2y^2+y^4 &=& (x^2+xy+y^2) (x^2-xy+y^2) \\ &=& x^4-x^3y+x^2y^2+x^3y-x^2y^2+xy^3+x^2y^2-xy^3+y^4 \\ &=& x^4+x^2y^2+y^4.\label{eg:wrongpf1} \end{eqnarray*}$

Here is the reason. When we place

$x^4+x^2y^2+y^4 = (x^2+xy+y^2) (x^2-xy+y^2)$

at the start of the proof, by convention, we are proclaiming that $$x^4+x^2y^2+y^4$$ is indeed equal to $$(x^2+xy+y^2) (x^2-xy+y^2)$$. However, this is what we are asked to prove. Before we have actually proved that it is true, we do not know yet, whether they are equal. Therefore, it is wrong to start the proof with it.

Example $$\PageIndex{3}$$

For the same reason, the following “proof” of the identity $x^3-y^3 = (x-y) (x^2+xy+y^2)$ is unacceptable:

$\begin{array}{lcl} x^3-y^3 &=& (x-y) (x^2+xy+y^2) \\ x^3-y^3 &=& x^3-x^2y+x^2y-xy^2+xy^2-y^3 \\ x^3-y^3 &=& x^3-y^3\label{eg:wrongpf2} \end{array}$

By putting $$x^3-y^3$$ on the left-hand side of every line, this becomes (by convention) a collection of three equations. In a nutshell, the argument starts with an equation and we simplify until we obtain something we know is true. If this format is valid, we can “prove” that $$21=6$$, as follows:

$\begin{eqnarray*} 21 &=& 6 \\ 6 &=& 21 \\ 27 &=& 27 \end{eqnarray*}$

By writing $$21=6$$ at the beginning of the proof, what we really say is “Assume $$21=6$$ is true.” But this is what we intend to prove. Thus, in effect, we are putting the cart in front of the horse, which is logically incorrect. There is another explanation why this proof is incorrect. We shall discuss it in Section 2.3.

In brief: we cannot start with the given identity and simplify both sides until we obtain an equality (or an equation of the form $$0=0$$).

Example $$\PageIndex{4}$$

Show that $$\frac{1}{6}\,k(k+1)(2k+1) + (k+1)^2 = \frac{1}{6} (k+1)(k+2)(2k+3)$$.

Solution 1

We can use the “meet in the middle” approach. Recall that we cannot simplify both sides simultaneously. Instead, we should expand the two sides separately, and then compare the results. We also suggest adding more writing (in words) to help with the explanation.

$\begin{eqnarray*} \textstyle \frac{1}{6}\,k(k+1)(2k+1) + (k+1)^2 &=& \textstyle \frac{1}{6}(2k^3+3k^2+k) + (k^2+2k+1) \\ &=& \textstyle \frac{1}{3}\,k^3+\frac{3}{2}\,k^2+\frac{13}{6}\,k+1. \label{eg:provingID-02}\end{eqnarray*}$
The right-hand side expands into
$\begin{eqnarray*} \textstyle \frac{1}{6} (k+1)(k+2)(2k+3) &=& \textstyle \frac{1}{6}(2k^3+9k^2+13k+6) \\ &=& \textstyle \frac{1}{3}\,k^3+\frac{3}{2}\,k^2+\frac{13}{6}\,k+1. \end{eqnarray*}$
Since both sides yield the same result, they must be equal.

Although the proof is correct, it requires two sets of computation. It is much easier to use either the left-to-right or the right-to-left approach.

Solution 2

A better alternative is to start from the left-hand side and simplify it until we obtain the right-hand side. Our secret weapon is factorization:

$\begin{eqnarray*} \textstyle \frac{1}{6}\,k(k+1)(2k+1) + (k+1)^2 &=& \textstyle \frac{1}{6} (k+1) [k(2k+1)+6(k+1)] \\ &=& \textstyle \frac{1}{6} (k+1)(2k^2+7k+6) \\ &=& \textstyle \frac{1}{6} (k+1)(k+2)(2k+3). \end{eqnarray*}$

This approach is usually better and safer, because no messy computation is involved.

Hands-on Exercise $$\PageIndex{1}$$

Show that $\label{he:provingID-01}\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}.$ Be sure to use one of the three methods we discussed above.

## Summary and Review

• There are only three ways to prove an identity: left to right, right to left, or meet in the middle.
• Never prove an identity by simplifying both sides simultaneously.

## Exercises $$\PageIndex{1}$$

Exercise $$\PageIndex{1}\label{ex:provingid-01}$$

Let $$x$$ and $$y$$ be any real numbers. Prove that $(x+y)^3 = x^3+3x^2y+3xy^2+y^3.$

Exercise $$\PageIndex{2}\label{ex:provingid-02}$$

Let $$x$$ and $$y$$ be any real numbers. Prove that $(a-b)^4 = a^4-4a^4b+6a^2b^2-4ab^3+b^4.$

Exercise $$\PageIndex{3}\label{ex:provingid-03}$$

Prove that, for any distinct real numbers $$x$$ and $$y$$, $\frac{x^3-y^3}{x-y} = x^2+xy+y^2.$

Exercise $$\PageIndex{4}\label{ex:provingid-04}$$

Prove that, for any integer $$k$$, $\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}.$

Exercise $$\PageIndex{5}\label{ex:provingid-05}$$

Prove that, for any integer $$k$$, $\frac{k^2(k+1)^2}{4} + (k+1)^3 = \frac{(k+1)^2(k+2)^2}{4}.$