To find the absolute value of a number is to find its magnitude, ignoring whether the number is positive or negative. The mathematical shorthand for absolute value is a pair of vertical lines enclosing a number or an expression.
So, $|-3.1|$|−3.1| means the magnitude of the number $-3.1$−3.1, which is $3.1$3.1. You can think of it as the distance from $0$0, to $-3.1$−3.1 on a number line. The distance is $3.1$3.1 units from zero. Notice how this is the same distance as $3.1$3.1 is from zero on a number line. So $|3.1|=|-3.1|=3.1$|3.1|=|−3.1|=3.1.
Evaluate | (a) $|-5|$|−5| |
(b) $|-5+9|$|−5+9| | |
(c) $|-5|+|9|$|−5|+|9| |
The absolute value of a negative number is its negative. So, $|-5|=-(-5)=5$|−5|=−(−5)=5.
We evaluate the sum $-5+9$−5+9 before applying the absolute value function. So, $|-5+9|=|4|=4$|−5+9|=|4|=4.
We evaluate separate absolute value terms before adding them. So, $|-5|+|9|=5+9=14$|−5|+|9|=5+9=14.
Note that $|-5+9|<|-5|+|9|$|−5+9|<|−5|+|9|.
We can show on a number line that if $|x|
The table shows that absolute values less than $4$4 only occur for numbers between $-4$−4 and $4$4.
The numbers $-4$−4 and $4$4 themselves are not included because it is not true that $|-4|<4$|−4|<4 and it is not true that $|4|<4$|4|<4. However, similar statements using $\le$≤ instead of $<$< would be true and the end-points would be included in the interval.
We can also visualise absolute values on a number line.
$|x|\le2$|x|≤2 means show me the values of $x$x, so that all the distances from $0$0 are less than or equal to $2$2. This we can see on the number line is this
$|x|<2$|x|<2 would be similar to above except this time we are strictly less than $2$2, so we use open points to mark the end of the interval.
$|x|>2$|x|>2 means we are now interested in all the values of $x$x that have a distance of more than $2$2 on the number line. This means we have $2$2 sections, all those points to the right of $2$2, and all the points to the left of $-2$−2.
$|x|\ge2$|x|≥2 means the same as above, except for now we want greater than or equal to a distance of $2$2, so we use a closed point to start the intervals.