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A 50-point math problem (like math)
There are many irrational numbers.

This is a problem of equivalence of finite sets, which is different from finite sets.

First of all, explain what "many" is. Rational number and irrational number are not equal, that is, one-to-one correspondence cannot be established. If two sets can establish a one-to-one correspondence, they are said to be equal (that is, "as many").

There are as many equivalent infinite sets as there are finite sets, which may be different intuitively. For example, integers and even numbers can correspond to each other (n corresponds to 2n), so they are equivalent.

Because rational numbers can be written in the form of integer fractions, rational numbers and integer pairs are equivalent; And because the integer pair (0,

0)、(0,

1)、( 1,

0)、( 1,

1) ... can be arranged in an ordered column (positive and negative can be staggered), so integer pairs and natural numbers are also equal.

Similarly, because the irrational numbers are1.1415926 …, 2.14/5926 …, 3.1415926. So irrational numbers will not be less than natural numbers, so they will not be less than rational numbers.

We just need to explain now that irrational numbers and natural numbers are not equal.

We use reduction to absurdity. Inverse irrational numbers can be lined up (so they can be numbered as 1, 2, 3…):

x.xxxx…

x.xxxx

……

We can find a new irrational number. Its first number is different from the first number in the above series, and its second number is different from the second number in the series ... So this new irrational number is not in the series, which is a contradiction. This contradiction shows that irrational numbers cannot be lined up, that is, irrational numbers are more than natural numbers, and therefore more than rational numbers.