Using the Identity and Inverse to Write Equivalent Expressions

Using the Identity and Inverse to Write Equivalent Expressions exercise

• Describe the additive and multiplicative inverses and identities.

  Hints

  Solve the following equation:

  $\begin{array}{rcl} x+5 & = & ~0\\ \color{#669900}{-5} & &\color{#669900}{-5}\\ x & = & ~-5 \end{array}$

  Solve the equation $8\times x=1$.

  Solution

  We are interested in the identity and inverse for addition and for multiplication:

  Let's start with addition:

  Which number can you add to any other number without changing the number? It's $0$, or the additive inverse.

  For example, $5+0=5$.

  In general, we have $x+0=x$ for any given $x$. This means that no matter which number you add zero to the sum is the number itself.

  The other way around we get the additive inverse: the number you have to add to a given number to get zero.

  We can see that $5-5=0$. We can write this as an addition of two numbers too: $5+(-5)=0$. So, $-5$ is the additive inverse of $5$.

  Generally $-x$ is the additive inverse for a given number $x$:

  • $x+(-x)=0$
  • $(-x)+x=0$

  Now, let's have a look at multiplication:

  The multiplicative identity is the number $1$. No matter which number you multiply by $1$ the product is the number itself. For example, $8\cdot 1=9$, or in general $x\cdot 1=x$.

  The other way around: the multiplicative inverse of a given number is the number you have to multiply the given number by to get $1$. Let's have a look at an example:

  $8\cdot \frac 18=1$.

  We can generalize it for a given non zero number $x$: $x\cdot \frac1x=1$.

• Determine which statements are true.

  Hints

  • The additive identity is $0$.
  • The multiplicative identity is $1$.

  Just look at the following examples:

  • $4+(-4)=0$
  • $7+0=7$
  • $8\times 1=1\times 8=8$
  • $7\div 7=1$

  Solution

  Here you have a few statements concerning the additive as well as multiplicative identity and inverse.

  If you add one rabbit to five rabbits the sum is five rabbits. We have that $5+1=6$. So, this statement is false.

  You can add zero to any given number without changing this number. Let any given number be represented by $x$, so you get $x+0=x$. This is true.

  The additive inverse of $5$ is $-5$. In general, the additive inverse of any given number $x$ is $-x$, so this statement is true.

  If you add any number to its additive inverse you get the multiplicative identity. Let any number be $x$. Then the additive inverse of $x$ is $-x$. Adding these two numbers together, we get $x+(-x)=0$. As $0$ is not the multiplicative identity, but rather the additive identity, this statement is false.

  If you add any number to its additive inverse you get the additive identity. For any number $x$, we have $x+(-x)=0$. This statement is true because $0$ is the additive identity, not the multiplicative one.

  Multiplying any number by zero leads to the number itself. This is a false statement. If you multiply any number by $0$ the product is $0$.

  If you divide any number by itself you get the multiplicative identity. Dividing any non zero number by itself can be written as $\frac xx=x\times \frac1x=1$. So, this statement is true.

• Decide which inverse or identity is being applied to $x$ in the equation.

  Hints

  The result of adding or multiplying by the inverse is the corresponding identity.

  • The additive identity is $0$.
  • The multiplicative identity is $1$.

  Here you see some examples:

  • $1+0=1$ ... for the additive identity
  • $2+(-2)=0$ ... for the additive inverse
  • $3\cdot 1=4$ ... for the multiplicative identity
  • $4\cdot \frac14=1$ ... for the multiplicative inverse

  Solution

  Letting $x$ be any number, we have that:

  • The additive identity is $0$, so $\rightarrow$ $x+0=x$.
  • The additive inverse of $x$ is $-x$, so $\rightarrow$ $x+(-x)=0$.
  • The multiplicative identity is $1$, so $\rightarrow$ $x\cdot 1=x$.
  • The multiplicative inverse of $x\neq 0$ is $\frac1x$, so $\rightarrow$ $x\cdot \frac1x=1$.

• Identify which inverse or identity $x$ represents for each equation.

  Hints

  Here is an example of using the additive identity: $7+0=7$.

  An example of using the additive inverse is $7+x=0$.

  Here, we have that $x=-7$.

  The solution to the equation $14\times x=14$ is the multiplicative identity $x=1$.

  Solution

  Keep the following in mind:

  • The additive identity is $0$ $\rightarrow$ $x+0=x$
  • The additive inverse of $x$ is $-x$ $\rightarrow$ $x+(-x)=0$
  • The multiplicative identity is $1$ $\rightarrow$ $x\times 1=x$
  • The multiplicative inverse of $x\neq 0$ is $\frac1x$ $\rightarrow$ $x\times \frac1x=1$

  So for the given equations, we have that $x$ represents the following inverses and identities:

  The additive identity:

  • $28+x=28$
  • $12+x=12$

  The additive inverse:

  • $5+x=0$
  • $x+11=0$

  The multiplicative identity:

  • $12\times x=12$
  • $2\times x=2$

  The multiplicative inverse:

  • $3\times x=1$
  • $5\times x=1$

• Find the right identity or inverse needed to make the equation true.

  Hints

  Write a fraction $\frac 12$ as follows $1\div 2$.

  The additive identity is $0$ and the multiplicative identity is $1$.

  For any given number $x$ the additive inverse is $-x$ and, if $x\neq 0$, the multiplicative inverse is $\frac 1x=1\div x$.

  Solution

  Keep the following in mind:

  • The additive identity is $0$.
  • The multiplicative identity is $1$.
  • The additive inverse of any given number $x$ is $-x$. Just change the sign.
  • The multiplicative inverse of any non zero number $x$ is $\frac1x=1\div x$, the reciprocal.

  With this we can quickly solve the following exercises:

  1. $5+(\mathbf{-5})=0$
  2. $3\times(\mathbf{1\div 3})=1$
  3. $4+(\mathbf{0})=4$
  4. $7\times(\mathbf{1})=7$
  5. $4+(\mathbf{-4})=0$
  6. $13\times(\mathbf{1\div 13})=1$

• Recall the definitions of additive identity, multiplicative identity, additive inverse, and multiplicative inverse.

  Hints

  Keep the following generalization in mind:

  • The additive identity is $0$ $\rightarrow$ $x+0=x$
  • The additive inverse of $x$ is $-x$ $\rightarrow$ $x+(-x)=0$

  Let's have a look at few examples:

  • $3+0=3$
  • $3+(-3)=0$
  • $3\times 1=3$
  • $3\times \frac13=1$

  Solution

  The additive identity is $0$ $\rightarrow$ $x+0=x$. You can add $0$ to any given number and the sum is always $x$ itself.

  The additive inverse of $x$ is $-x$ $\rightarrow$ $x+(-x)=0$.

  The multiplicative identity is $1$ $\rightarrow$ $x\times 1=x$. You can multiply any given number by $1$ and the product is the number itself.

  The multiplicative inverse of $x\neq 0$ is $\frac1x$ $\rightarrow$ $x\times \frac1x=1$.