Proof That ##\sqrt{x}## Isn't Rational (Unless ##x## is a Perfect Square)

[marino]

According to you this theorem is correct?

Exercise 1.2 * Proof that $\sqrt{x}$ isn't a rational number if $x$ isn't a perfect square (i.e. if $x=n^2$ for some $n∈ℕ$).

In effect, if $x=\frac{25}{9}$, so $x$ isn't a perfect square, then $\sqrt{x}=\sqrt{\frac{25}{9}}=\frac{5}{3}$ is rational.

 

[jedishrfu]

But for this proof to work:

x must be a member of N too

and so sqrt of x is not a rational number.

@fresh_42 can answer this better.

 

[fresh_42]

There is nothing to add. Either restrict $x\in \mathbb{N}$ or generalize $n \in \mathbb{Q}$.

 

[PeroK]

According to you this theorem is correct?

Exercise 1.2 * Proof that $\sqrt{x}$ isn't a rational number if $x$ isn't a perfect square (i.e. if $x=n^2$ for some $n∈ℕ$).

In effect, if $x=\frac{25}{9}$, so $x$ isn't a perfect square, then $\sqrt{x}=\sqrt{\frac{25}{9}}=\frac{5}{3}$ is rational.

To put it another way.

If $x$ is a positive integer, then either $\sqrt{x}$ is a positive integer; or, $\sqrt{x}$ is irrational.

 

[WWGD]

One way of doing the proof for Rationals is to generalize the argument that $\sqrt 2$ is not Rational to $\sqrt n$.

 

[Mark44]

In effect, if $x=\frac{25}{9}$, so $x$ isn't a perfect square

But here x is a perfect square ($(\frac 5 3)^2 = \frac {25} 9$, but as has already been pointed out, x is not a positive integer.

 

[WWGD]

But here x is a perfect square ($(\frac 5 3)^2 = \frac {25} 9$, but as has already been pointed out, x is not a positive integer.

But you need to be careful on how you define a perfect square otherwise every no negative real number is a perfect square.

 

[Mark44]

But you need to be careful on how you define a perfect square otherwise every no negative real number is a perfect square.

Understood, but when we talk about perfect squares, the context is seldom real numbers. In this thread the context was $x \in \mathbb N$ (more specifically, from post #1, $x = n^2$ with $n \in \mathbb N$) and the OP gave an example with $n^2 = \frac {25} 9$.