Question

\sqrt{3},\sqrt{5},\sqrt{6} докажите что они являются иррационалами! ( Помогите, плиз)

Answer 1

Иррациональное число может быть представлено в виде бесконечной непериодической десятичной дроби.

√3=1,732050807568877....

√5=2,23606797749979...

√6=2,449489742783178...

Answer 2

Irrationality of √3, √5, and √6

To prove that √3, √5, and √6 are irrational numbers, we need to show that they cannot be expressed as the ratio of two integers. We can do this by assuming the opposite, that is, assuming that √3, √5, and √6 are rational numbers, and then deriving a contradiction.

Let's start with √3. We assume that √3 is rational, which means it can be expressed as the ratio of two integers, p and q, where q is not equal to 0 and p and q have no common factors other than 1. So, we have:

√3 = p/q

Squaring both sides of the equation, we get:

3 = (p^2)/(q^2)

Rearranging the equation, we have:

p^2 = 3q^2

This implies that p^2 is divisible by 3. From this, we can conclude that p itself must be divisible by 3. Let's say p = 3k, where k is an integer. Substituting this back into the equation, we get:

(3k)^2 = 3q^2

9k^2 = 3q^2

Dividing both sides by 3, we have:

3k^2 = q^2

This implies that q^2 is divisible by 3, and therefore q itself must be divisible by 3. However, this contradicts our initial assumption that p and q have no common factors other than 1. Hence, our assumption that √3 is rational is false, and therefore, √3 is an irrational number.

We can follow a similar approach to prove that √5 and √6 are also irrational numbers. By assuming that they are rational and following the same steps, we will eventually arrive at a contradiction, proving that √5 and √6 are irrational.

In conclusion, √3, √5, and √6 are irrational numbers because they cannot be expressed as the ratio of two integers.