Algebra I

When 1=2

The most common source of mistakes in algebra is making a change to an equation or expression that isn't allowed. Apart from being perfect calculating machines, how can we avoid these mistakes?

\[ \begin{array} {c r c l }
\color{red} {1} & ab & = & a^2 \\ \color{orange} {2} & ab - b^2 & = & a^2 -b^2 \\ \color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\ \color{green} {4} & b & = & a+b \\ \color{blue} {5} & b & = & b+b \\ \color{indigo} {6} & b & = & 2b \\ \color{purple} {7} & 1 & = & 2. \end{array} \]

We regularly use our intuition about numbers to prevent obvious errors in arithmetic. For example, we don't have to work hard to see that \(249+356=95\) is wrong. We know that two positive three digit numbers could not possibly sum to something less than 100!

In the following questions, we'll look at a common algebraic mistake and try to train ourselves to recognize it in many circumstances.

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

This course will dive into the reasons that algebra steps work in the first place. The goal is to make algebra less mysterious and show there is something flexible and interesting behind the rules.

For example, 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 to perform the cancel? \((\)They're expressed in the form of a newly defined number \(Q\) where \(Q \neq 0.)\)

           
×

Problem Loading...

Note Loading...

Set Loading...