Look at a and 1

Try this

Start with $(a + 1)^{2} = a^{2} + 2 a + 1$ . Put in $3$ , $- 2$ and $\frac{1}{2}$ one at a time, then $x + 1$ and $2 x$ . Does it ever fail?
Switch to $2 a + 1 = 7$ . Of the seven fillings, how many make it true?
Switch to $a^{2} = a$ . Which numbers make it true this time?
Switch to $\frac{a}{a} = 1$ and put in $0$ . What does the machine say?
Last, $a + 1 = a$ : can you find anything that makes it true? Once you have tried three fillings for a statement, click “So when does it hold?” and see if you were right.

The five statements look alike, but they behave completely differently. One that is true whatever you put in is called an identity: it describes a rule of arithmetic itself. One that is true only for particular numbers is called an equation, and those numbers are its solutions. This whole volume is about finding those numbers.

Predict first

$a$ is a number. Must $- a$ be negative?

Yes, always negativeNot necessarilyNever negative

Choose an answer first; then you can see the result.

Why

Why does the first statement work whatever you put in? Because it was derived step by step from the laws of arithmetic, and those laws hold for every number.

Why (a+1)² = a²+2a+1 always holds

Step 1 / 6

Put in a number
Put $3$ in the slot: the left side is $4^{2} = 16$ and the right side is $9 + 6 + 1 = 16$ . Equal.

$(3 + 1)^{2} = 3^{2} + 2 \cdot 3 + 1$
Put in a negative number
Put in an expression
The reason: only the laws of arithmetic
Compare: an equation
Change the number and it fails

The hidden 1

Lots of places hide a $1$ that is not written:

a = 1 \cdot a = a^{1} = \frac{a}{1}, - a = (- 1) \cdot a

When you collect like terms or take out a common factor, bring it out into the open:

$3 x - x = 3 x - 1 x = 2 x$
$ab + a = a \cdot b + a \cdot 1 = a (b + 1)$

The other way round, $a$ can also be treated as a single block. In the identity above, replace $a$ with $x + 1$ , with $2 x$ , or with something even longer, and it still holds. You will use this trick all the time when solving equations: treat a big chunk as a single letter.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $\frac{a}{a} = 1$

Put in only numbers and expressions that are not $0$ : every one of them works.

Practice

Exercise 1Basic

Find the value of $a^{2} - 2 a + 1$ when $a = - 3$ .

Exercise 2Basic

Collect like terms: $5 x - x + 2 x$ .

Exercise 3Advanced

Which numbers $a$ make $a^{2} = 2 a$ true? (List all of them.)

Exercise 4Challenge

First check that $(a + 2)^{2} - (a - 2)^{2} = 8 a$ is an identity, then use it to work out $100 2^{2} - 99 8^{2}$ in your head.

Try this

Why

Why (a+1)² = a²+2a+1 always holds

The hidden 1

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Seeing a minus sign and calling it negative

Forgetting the hidden 1 in front of x

Cancelling without checking the condition