As shown in the graph above, it's called a consequence of the Pythagorean theorem: The length of the diagonal of a square is the square root of two times its side. The length of the square is a rational number while the length of diagonal is irrational. While we can just keep getting the 1/2 of the existing square to get infinitely many rational numbers, these natural numbers are countably many. But the number of irrationals are far more than the rationals, which is illustrated by the cardinality of the rationals and irrationals. |R\Q| >|Q|.
Now here is a trick to show why there are more irrationals than rationals.
We do this by trying to list all the real numbers in the interval (0,1).
First, because each of the rational numbers can be associated with a natural number by listing there like below.
In this way, each number is followed by its negative version, except zero, which doesn't have a different negative version. he fractions, if considered as improper fractions, are ordered by the sum of the numerator and denominator, starting with the fraction with the largest numerator: thus, we go through 4/1, 3/2, 2/3,1/5 in order. (Earlier, 2/2 was omitted, since that equals 1, which is already on the list.)
N Q
1 1/1
2 -1
3 2/1
4 -2/1
5 1/2
6 -1/2
7 3
8 -3
9 1/3
10 -1/3
11 4
12 -4
... ...
Then we can use all the natural numbers to list all the rational numbers. We have |Q| = |N|
Now we are trying to show there are more irrational number than rationals by trying to list all the irrationals by natural numbers.
.27231364585...
.40861132064...
.01678396536...
.81171944218...
.82122547689...
.17218943291...
...
We are trying to list all the real numbers by associating each with a natural number. But it seems there are always at least 1 more irrational number than all the rational numbers that we can list. Assume we listed all the natural number with each there's a corresponding irrational number, we can get a new irrational numbers by picking up the irrational on the diagonal of the existing list.
Because the new number is attained by picking the jth digit of the decimal on jth row, so it's different from any of the existing irrationals.
Therefore there are more irrational numbers than natural numbers, thus there are more irrational numbers than rationals.
I think you have to construct the new number by *changing* the digits you have in bold. Then the new number differs from the first number in the first decimal place, the second number in the second decimal place, and so on. I.e., it differs from all numbers in the list in at least one decimal place. I also think you have to capitalize on being able to add more decimal places.
ReplyDeleteI find this whole notion troubling, because it seems to me that for any irrational number that you put in my hand I can construct a rational number. Let the irrational number be given in the form N.M, where M has D decimal places.
ReplyDeleteOk, fine. [ (10^D)*N + M ] / (10^D) is a rational number with the same value. D may rise toward infinity, but there is still a rational number to match every single irrational one.