Choose a number you know for sure is rational. How can you use it to get other rational and irrational numbers? Is it possible to apply the same process for any interval?
Example Interval I: 1.41≤ x ≤ sqrt(2) Example Interval II: 0.71≤ x ≤ sqrt(2)/2 Example Interval III: 0.47 ≤ x ≤ sqrt(2)/3 Does Every Interval Contain Rational and Irrational Numbers? Yes, see solution.
Practice makes perfect
We can find infinitely many intervals satisfying the given conditions. We will only mention three possibilities.
Finding 3 Example Intervals
Let's start by choosing one irrational number. For simplicity, let's choose sqrt(2).
sqrt(2)=1.414213 ...
We will create three intervals using this number.
Interval I
If we round this number to its second decimal place, we get 1.41. Since it only has two decimal digits, 1.41 is a rational number. Let's define our first interval as all numbers greater than or equal to 1.41 and less than or equal to sqrt(2).
1.41≤ x ≤ sqrt(2)
The above interval contains the irrational number sqrt(2) and the rational number 1.41.
Interval II
Since sqrt(2) is irrational, the number sqrt(2)2 is irrational as well. This is because it is not the ratio between two integers.
l Not integer→ Integer→ sqrt(2)/2=0.707106...
By rounding this new number to its second decimal place, we obtain 0.71. Let's define our second interval as all numbers greater than or equal to 0.71 and less than or equal to sqrt(2)2.
0.71≤ x ≤ sqrt(2)/2
The interval above contains the irrational number sqrt(2)2 and the rational number 0.71.
Interval III
Similarly, sqrt(2)3 is guaranteed to be an irrational number.
sqrt(2)/3 =0.471404...
If we round this number to the nearest hundredth, we obtain 0.47. Let's define our third interval as all numbers greater than or equal to 0.47 and less than or equal to sqrt(2)3.
0.47 ≤ x ≤ sqrt(2)3
The interval above contains the irrational number sqrt(2)3 and the rational number 0.47. We could repeat the same process an infinite number of times, finding as many intervals as we want.
Does Any Given Interval Contain Rational and Irrational Numbers?
We want to see whether any interval contains rational and irrational numbers. To do so, we will consider an arbitrary interval.
3 ≤ x ≤ 5
Since the above interval includes 3 and 5, we are sure it contains rational numbers. Moreover, the irrational number π is also contained in this interval. If we try, we can find other irrationals included in the interval. For example, the number sqrt(15).
Ď€&=3.141592...
sqrt(15)&=3.872983 ...
Let's show the interval and these numbers on the real number line.
For every irrational number not contained in this interval, we can find an irrational number included in the interval. We do so by dividing it by some integer. Let's consider the irrational numbers sqrt(67) and sqrt(187).
sqrt(67)& = 8.185352 ...
sqrt(187)& =13.674794...
To find corresponding irrational numbers included in the interval, we can divide sqrt(67) by 2 and sqrt(187) by 3.
sqrt(67)/2 & = 4.092676 ... [0.8em]
sqrt(187)/3 & = 4.558264...
We have found two more irrational number in our interval.
We could continue this process indefinitely, finding as many irrational numbers contained in our interval as we want. On the other hand, we can do something similar but using integers to obtain rational numbers included in the interval.
7/2 = 3.5,
8/2 = 4, and
19/4= 4.75
Let's show these numbers in the interval.
This interval was chosen arbitrarily and is by no means special. We could have followed exactly the same process with any interval. Therefore, not only will any given interval contain rational and irrational numbers, but it will also contain infinitely many of them!
