Question

Prove that V35 is irrational number​

Answer 1

Answer:

n mathematics, the irrational numbers are all the real numbers which are not rational numbers, the latter being the numbers constructed from ratios (or fractions) of integers. When the ratio of lengths of two line segments is an irrational number, the line segments are also described as being incommensurable, meaning that they share no "measure" in common, that is, there is no length ("the measure"), no matter how short, that could be used to express the lengths of both of the two given segments as integer multiples of itself.

Among irrational numbers are the ratio π of a circle's circumference to its diameter, Euler's number e, the golden ratio φ, and the square root of two;[1][2][3] in fact all square roots of natural numbers, other than of perfect squares, are irrational.

It can be shown that irrational numbers, when expressed in a positional numeral system (e.g. as decimal numbers, or with any other natural basis), do not terminate, nor do they repeat, i.e., do not contain a subsequence of digits, the repetition of which makes up the tail of the representation. For example, the decimal representation of the number π starts with 3.14159, but no finite number of digits can represent π exactly, nor does it repeat. The proof that the decimal expansion of a rational number must terminate or repeat is distinct from the proof that a decimal expansion that terminates or repeats must be a rational number, and although elementary and not lengthy, both proofs take some work. Mathematicians do not generally take "terminating or repeating" to be the definition of the concept of rational number.

Irrational numbers may also be dealt with via non-terminating continued fractions.

As a consequence of Cantor's proof that the real numbers are uncountable and the rationals countable, it follows that almost all real numbers are irrational.[4]

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