Identity Properties of Addition and Multiplication

The identity property of addition is that given some number \(Q\), \( Q + 0 = Q .\) The identity property of multiplication is that given some number \(Z,\) \( Z \times 1 = Z .\)

Identity properties are fundamental to the workings of traditional arithmetic, and in many related systems. They are essentially "operations that do nothing", that leave the identity of a number untouched. This is useful in algebraic manipulation.

Note these properties can apply to number systems other than the real numbers! They are phrased this way here for simplicity.

Identity Property of Addition for Real Numbers: Given any real number \(Q,\) if 0 is added to \(Q,\) the result is \(Q.\) That is \( Q + 0 = 0 + Q = Q .\)

Identity Property of Multiplication for Real Numbers: Given any real number \(Z,\) if 1 is multiplied by \(Z,\) the result is \(Z.\) That is, \( Z \times 1 = 1 \times Z = Z .\)

This particular identity shows up in mental arithmetic. If we have the sum

\[ 122 + 59 \]

we can make the arithmetic easier to do by first "adjusting" the 59 so the last digit is 0. Note if we just add 1 we can get 60. However, adding 1 arbitrarily will result in a different number! What we can do instead is add 1 and subtract 1; this is equivalent to adding 0. By applying an identity property, we are guaranteed the number doesn't change.

\[ 122 + 59 + 1 - 1 = 122 + 60 - 1 \]

Thus, mentally, we can just 1.) add 122 to 60 and then 2.) adjust by subtracting by 1, which is easier than mentally adding the original ones digits of 2 and 9 and worrying about a carry digit.

Here's an algebra example: suppose you have the expression

\[ x^2 + 6x - 1 \]

and you want to do completing the square. This will rewrite the expression in the format \( (x-h)^2 + k \) (and makes it easy, for example, to find the vertex of a parabola).

If we think of the expression in the format \( ax^2 + bx + c ,\) we can take half of the \(b\) term and square it; the result can be combined in the expression to complete the square. With the example above, \( \frac{1}{2} \times 6 = 3 ,\) and \(3^2 =9,\) so our goal is for the expression to contain \( x^2 + 6x + 9 .\) However, we can't just change the -1 to a 9; what we can do is add 9 and subtract 9, which is equivalent to adding 0. Since adding 0 is an identity property, nothing is changed.

\[ x^2 + 6x + 9 - 9 - 1 \]

Now we can finish completing the square, and combine \( -9 - 1 \) as a separate term.

\[ (x+3)^2 - 10 \]

If we are dealing with an equation, note an alternative to using the addition property of identity would be to add the same value to both sides of the equal sign. For example, given \( 3 = x^2 + 6x - 1 ,\) we could add 9 to both sides and get \( 3 + 9 = x^2 + 6x + 9 - 1 .\) However, sometimes the expression being worked with is embedded in a larger equation and using both sides of the equal sign isn't possible; in such a scenario the addition property of identity is completely necessary.

No problem found with slug "a-b-c"

Simplifying and reducing fractions often applies this identity. For example, suppose you have the fraction \( \frac{4}{25} \) but you want a denominator of 100 (perhaps in making a percentage). You want to multiply the 25 by 4, turning it into 100. However, you can't just arbitrarily do that, because it changes the number. What you can do is multiply by \( \frac{4}{4} ,\) which is equivalent to multiplying by 1. The identity property of multiplication means that this won't change the number:

\[ \frac{4}{25} \times \frac{4}{4} = \frac{16}{100} .\]

The algebraic analogue of fractions is rational expressions, where the same sort of logic applies.

\[ \frac{2x}{3x^2} + \frac{x}{3x} \]

To do the addition with the example above, we'd like to have both denominators read \( 3x^2 .\) However, the second term \( \frac{x}{3x} \) is missing an \(x\) in the denominator. We can't just multiply it in without changing the number, but we can apply the identity property of multiplication and use \( \frac{x}{x} \) instead.

(Warning note: \( \frac{x}{x} \) is not 1 if \(x\) is 0. Since the current expression already is invalid when \(x=0,\) this does not change the situation here.)

\[ \frac{2x}{3x^2} + \frac{x}{3x} \frac{x}{x} = \frac{2x}{3x^2} + \frac{x^2}{3x^2} \]

Now that the denominators are the same, we can add and get \( \frac{2x+x^2}{3x^2} .\)