B Perfect square

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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.
 
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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

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

PeroK

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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.
 
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WWGD

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

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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.
 
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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##.
 

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