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They don't stop.
Suppose an irrational number can be written precisely in decimal form, with n digits after the decimal point. Then if you multiply the decimal value by 10n you will get an integer, say k. Then the decimal representation is equivalent to k/10n, which is a ratio of two integers and so the number, by definition, is rational - not irrational.
But irrational numbers are decimals that can't be expressed as fractions
They don't stop.
They don't stop.
They don't stop.
Suppose an irrational number can be written precisely in decimal form, with n digits after the decimal point. Then if you multiply the decimal value by 10n you will get an integer, say k. Then the decimal representation is equivalent to k/10n, which is a ratio of two integers and so the number, by definition, is rational - not irrational.
But irrational numbers are decimals that can't be expressed as fractions
They don't stop.
They don't stop.
Yes, However, it is not defined that way. It is defined as a number that cannot be expressed precisely as a ratio of two real numbers (a fraction). But that is equivalent to a non-repeating decimal.
Irrational numbers are precisely those real numbers that cannot be represented as terminating or repeating decimals. Log 216 = 2.334453751 terminates and is therefore not irrational.
An irrational number by definition can not be exactly represented by a decimal that terminates or recurs. The moment a decimal terminates, or settles into a repeating pattern, it is rational.
No because irrational numbers can't be expressed as fractions
Because rational numbers aren't able to be notated precisely in decimal form. They don't stop.
None of them: they are all rational since they can all be represented as terminating decimal numbers.
Yes, except that all irrational numbers will be non-terminating, non-repeating decimals.