Prove that 5^(2/3) is irrational

  • Thread starter P-Illiterate
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    Irrational
En resumen, se está tratando de demostrar que 5^(2/3) es irracional y se ha intentado escribir una prueba, pero no se ha llegado a ningún resultado. Se propone usar el hecho de que 52/3 es igual a la raíz cúbica de 25 para simplificar el proceso.
  • #1
P-Illiterate
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Homework Statement


Prove that 5^(2/3) is irrational

Homework Equations





The Attempt at a Solution



I tried writing a proof but that is not getting me any where.

This is what I did so far -

Show that 52/3 is irrational

Proof: Suppose that 52/3 is rational:
52/3 = a/b
52/33/2 = a3/2/b3/2
5(b3/2) = a3/2
Substitute a = 2n + 1, b = 2m + 1;

I don't know if I'm suppose to do like this?
 
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  • #2
P-Illiterate said:

Homework Statement


Prove that 5^(2/3) is irrational

Homework Equations





The Attempt at a Solution



I tried writing a proof but that is not getting me any where.

This is what I did so far -

Show that 52/3 is irrational

Proof: Suppose that 52/3 is rational:
52/3 = a/b
52/3 3/2 = a3/2/b3/2
5(b3/2) = a3/2
Substitute a = 2n + 1, b = 2m + 1;

I don't know if I'm suppose to do like this?

It might be simpler to use the fact that 52/3 = ##\sqrt[3]{25}##
 
  • #3
Mark44 said:
It might be simpler to use the fact that 52/3 = ##\sqrt[3]{25}##

:approve:
Gracias
 
  • #4
De nada
 

1. What does it mean for a number to be irrational?

A number is considered irrational if it cannot be expressed as a ratio of two integers (whole numbers). This means that it cannot be written in the form of a fraction with a non-zero denominator.

2. How can we prove that 5^(2/3) is irrational?

We can prove that 5^(2/3) is irrational using the proof by contradiction method. We assume that 5^(2/3) is rational and can be expressed as a fraction a/b, where a and b are integers with no common factors. Then we raise both sides to the power of 3, giving us 5^2 = (a/b)^3. Simplifying this equation, we get 25 = a^3/b^3. However, this means that 25 is a perfect cube, which is not possible since 25 is not a perfect cube. This contradicts our initial assumption, therefore proving that 5^(2/3) is irrational.

3. Can we use a different method to prove that 5^(2/3) is irrational?

Yes, we can also use the decimal representation method to prove that 5^(2/3) is irrational. The decimal representation of 5^(2/3) is approximately 2.924017738212866, and it continues infinitely without repeating. This shows that 5^(2/3) cannot be expressed as a terminating or repeating decimal, which is a characteristic of irrational numbers.

4. Is 5^(2/3) the only number that is irrational?

No, there are infinitely many irrational numbers. Some well-known examples include pi (π), the square root of 2 (√2), and the golden ratio (φ). In fact, the majority of real numbers are irrational, and only a small fraction can be expressed as rational numbers.

5. Why do we need to prove that 5^(2/3) is irrational?

Proving that 5^(2/3) is irrational is important because it helps us understand and classify different types of numbers. It also allows us to solve certain mathematical problems and equations that involve irrational numbers. Additionally, it helps us appreciate the beauty and complexity of mathematics.

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