Algebra I

When 1 = 2

Let \( a = b \).

Then, \[ \begin{array} {c r c l }
\color{red} {1} & ab & = & a^2 \\ \color{orange} {2} & ab - b^2 & = & a^2 -b^2 \\ \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \color{green} {4} & b & = & a+b \\ \color{blue} {5} & b & = & b+b \\ \color{indigo} {6} & b & = & 2b \\ \color{purple} {7} & 1 & = & 2. \end{array} \] 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: \[ \begin{array} {c r c l }
\color{red} {1} & ab & = & a^2 \\ \color{orange} {2} & ab - b^2 & = & a^2 -b^2. \end{array} \] \(b^2\) is subtracted from both sides of the equal sign. Is it equally valid to do the operation below instead? \[ \begin{array} {c r c l }
\color{red} {1} & ab & = & a^2 \\ \color{orange} {2} & b^2 - ab & = & b^2 - a^2 \\ \end{array} \]

When 1 = 2

\[ \begin{array} {c r c l }
\color{orange} {2} & ab - b^2 & = & a^2 -b^2 \\ \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \end{array} \]

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

\[ \begin{array} {c r c l }
\color{orange} {2} & ab - b^2 & = & a^2 -b^2 \\ \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \end{array} \]

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: \[ \begin{array} {c r c l }
\color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \color{green} {4} & b & = & a+b. \end{array} \] \( (a-b) \) 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? \((\)Assume \(Q \neq 0.)\)

         
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