How to locate some irrational numbers on the number line?

How to locate some irrational numbers on the number line?How to locate some irrational numbers on the number line?
The square root of 2 is irrational, the square root of 3, 5, 6, 7, 8, 10, and any other positive number that is not a perfect square. You can find more by dividing an integer by an irrational number. (5 / sqr rt of 2...)





There are more irrational numbers than there are rational numbers. There are aleph-null rational numbers and aleph-one irrational numbers. (Aleph-null is the smallest size of infinity, aleph-one is the second size.)How to locate some irrational numbers on the number line?
You can do it using geometrical constructions. Say you want to locate sqrt(2), make a right isoceles triangle having sides containing the right angle = 1 unit. Than the hypotenuse is sqrt(2). If you do this construction such that hypotenuse lies on the number line. You have located sqrt(2).





Similarly, for each rational number you can do some constructions and fins them out on the number line.
irrational numbers cannot be precisely located on a number line, they can only be approximated.
蟺 is between 3 and 4


e is between 2 and 3


sqrt(2) is between 1 and 2
Throw a dart, you will likely hit one. Seriously. Irrationals far and away are predominant.





Here is an easy method to find your own:





Take any rational number. Multiply, divide, add, or subtract it by pi, or e, or sqrt(2), etc, and you will have your very own irrational number.
I'm not sure how but the easiest way would be to check by using calculator
Irrational numbers are fractions or mixed fractions. The complete part of mixed fraction can be easily marked on number line. Now next unit slot should be divided in equal numer of parts equal to denominator of the fraction and mark parts equal to the numerator.


e.g.,


a b/c


Mark a on number line as usual. The part between a %26amp; a+1 be divided into c equal parts %26amp; take b parts of it to mark a b/c.
* Assume that \sqrt{2} is a rational number. This would mean that there exist integers m and n such that m/n = \sqrt{2}.


* Then \sqrt{2} = \frac{\sqrt{2}\cdot n(\sqrt{2}-1)}{n(\sqrt{2}-1)} = \frac{2n-\sqrt{2}n}{\sqrt{2}n-n} = \frac{2n-m}{m-n}.


* Since \sqrt{2} %26gt; 1, it follows that m %26gt; n\,, and it can be shown that m %26gt; 2n - m\,


got it?

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