Algebra I

Let $a = b$.

Then, $\begin{array} {c r c l } \color{#D61F06} {1} & ab & = & a^2 \\ \color{#EC7300} {2} & ab - b^2 & = & a^2 -b^2 \\ \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \color{#20A900} {4} & b & = & a+b \\ \color{#3D99F6} {5} & b & = & b+b \\ \color{#302B94} {6} & b & = & 2b \\ \color{#69047E} {7} & 1 & = & 2. \end{array}$ Which line is the first that contains an error in this proof?

When 1 = 2

Let's consider the transition from line 1 to 2: $\begin{array} {c r c l } \color{#D61F06} {1} & ab & = & a^2 \\ \color{#EC7300} {2} & ab - b^2 & = & a^2 -b^2. \end{array}$ $b^2$ is subtracted from both sides of the equal sign. Is it equally valid to do the operation below instead? $\begin{array} {c r c l } \color{#D61F06} {1} & ab & = & a^2 \\ \color{#EC7300} {2} & b^2 - ab & = & b^2 - a^2 \\ \end{array}$

When 1 = 2

$\begin{array} {c r c l } \color{#EC7300} {2} & ab - b^2 & = & a^2 -b^2 \\ \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \end{array}$

On the left-hand side of line 3, the value $b$ is "factored out" from $ab-b^2$ to get $b(a-b) .$

Is the equation below true? $ab - b^2 - a^2 = ab(1-b-a)$

When 1 = 2

$\begin{array} {c r c l } \color{#EC7300} {2} & ab - b^2 & = & a^2 -b^2 \\ \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \end{array}$

Changing $a^2 - b^2$ to $(a+b)(a-b)$ is known as the difference of squares identity (it was first introduced in a previous course, as shown in the video below). What if we started with fourth powers instead, that is, $a^4 - b^4 ?$ Which expression is equivalent?

When 1 = 2

Let's go back to the moment of mistake: $\begin{array} {c r c l } \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \color{#20A900} {4} & b & = & a+b. \end{array}$ $(a-b)$ does what is known as a "cancel" on both sides of the equal sign. Since the problem set up $a = b,$ this implies division by zero. Suppose that $a \neq b$ instead; what identity would be used to cancel? $($Assume $Q \neq 0.)$

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