Numbers: Real numbers
Ordering of real numbers
The real numbers together form a line on which an order is indicated. This means that, for each pair of numbers, we can determine which one is the larger, and which one is the smaller. Below a part of this line is shown, indicating the integers between #-4# and #4#, the rational number #-\frac{3}{2}# and irrational numbers #\sqrt{2}#, #\pi#, and #-\e#. The number #\pi# is the area of a circle with radius #1# and #\e# is Euler's number, for which we will give definition later on.
The ordering of the real numbers is determined by the fact that a number is smaller than another if it is located on the left of it of the number line. We can help to notate this using the symbol #\lt# (pronounced less than), which we write between two numbers to indicate that the left number is smaller than the right one. For example #-1\lt 0# and #\pi\lt 4#. We can also write a series of inequalities, such as
\[-1\lt 0\lt\ 1\lt \sqrt{2}\lt 2\lt \pi\lt 4\lt \frac{9}{2}\lt 5\]
The other symbols that we use to describe the ordening are all described in terms of #\lt# :
| symbol | statement | definition |
| #\le# | less than or equal to | #x\le y# if and only if #x\lt y# or #x=y# |
| #\gt# | greater than | #x\gt y# if and only if #y\lt x# |
| #\ge# | greater than or equal to | #x\ge y# if and only if #x\gt y# or #x=y# |
The four symbols #\lt#, #\le#, #\gt#, #\ge# are called inequality signs.
The inequality #\sqrt{2} \gt 1# means the same as #1\lt \sqrt{2}#.
The inequalities #\frac{1}{3}\ge 0.333# and #\frac{1}{3}\gt 0.333# are both true because #\frac{1}{3}# is not equal to #0.333#.
Below are the main rules of calculation for inequalities
Rules of calculation for inequalities
Let #x#, #y#, and #z# be real numbers. Then the following rules apply:
- Exactly one of the three statements "#x\gt y#", "#x\lt y#", " #x=y# "is true
- If #x\lt y# and #y\lt z#, then #x\lt z#
- If #x\gt y#, then #x+z\gt y+z#
- If #x\gt y#, then #-x\lt -y#
- If #x\gt y# and #z\gt0#, then #z\cdot x\gt z\cdot y#
- If #x\gt0# and #y\gt 0# then #x\cdot y \gt 0#
- If #x\gt0#, then #\frac{1}{x}\gt 0 #
- If #x\gt y\ge0#, then #x^2\gt y^2# and #\sqrt{x}\gt\sqrt{y}#
Proof of Rule 8: Assume #x\gt y\ge 0# . Because of rule 1 (in case #y\gt0#) holds #x\gt0#. Applying rule 5 twice we find #x^2=x\cdot x\gt x\cdot y\gt y\cdot y = y^2#. Applying rule 1 again gives #x^2\gt y^2#. This provides the first statement of Rule 8.
Next if #\sqrt{x}\lt \sqrt{y}#, then #x\lt y# results from the rule above, which is a contradiction with the hypothesis #x\gt y\ge0#.
Hence #\sqrt{x}\gt \sqrt{y}# has to be satisfied. We have seen that from #x\gt y\ge 0# it follows that #\sqrt{x}\gt\sqrt{y}#. This is the second statements of rule 8.
Because, #{9}\cdot 13 = 117 \lt 144 = 18 \cdot 8#, hence according to the rule invariance under multiplication by a positive number :
\[\frac{9}{8} =\frac{9\cdot 13 }{8\cdot 13} = \frac{1 }{8\cdot 13} \cdot 117 \lt \frac{1 }{8\cdot 13}\cdot 144 = \frac{18\cdot 8 }{8\cdot 13} =\frac{18}{13}\tiny.\]
Or visit omptest.org if jou are taking an OMPT exam.
