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

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Discussion Overview

The discussion revolves around the theorem asserting that ##\sqrt{x}## is not a rational number unless ##x## is a perfect square. Participants explore the conditions under which this theorem holds, particularly focusing on the nature of ##x## and its implications for rationality.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that for the theorem to hold, ##x## must be a member of the natural numbers, suggesting that if ##x## is not a perfect square, then ##\sqrt{x}## cannot be rational.
  • Others propose that the proof could be generalized to include rational numbers, indicating that the argument used for ##\sqrt{2}## could be extended to ##\sqrt{n}## for rational ##n##.
  • One participant points out that if ##x = \frac{25}{9}##, then ##\sqrt{x} = \frac{5}{3}##, which is rational, challenging the initial claim about perfect squares.
  • Another participant emphasizes the need for careful definitions of what constitutes a perfect square, warning against the assumption that all non-negative real numbers could be classified as such.
  • There is a reiteration that the context of the discussion primarily concerns positive integers, as indicated by the original post's reference to ##n^2## with ##n \in \mathbb{N}##.

Areas of Agreement / Disagreement

Participants express disagreement regarding the conditions under which the theorem applies, particularly about the definition of perfect squares and the inclusion of rational numbers. No consensus is reached on the validity of the theorem as stated.

Contextual Notes

Limitations include the ambiguity in the definition of perfect squares and the assumptions about the set from which ##x## is drawn. The discussion highlights the need for clarity on whether ##x## is restricted to natural numbers or can include rational numbers.

marino
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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.
 
There is nothing to add. Either restrict ##x\in \mathbb{N}## or generalize ##n \in \mathbb{Q}##.
 
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marino said:
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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One way of doing the proof for Rationals is to generalize the argument that ##\sqrt 2## is not Rational to ##\sqrt n##.
 
marino said:
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.
 
Mark44 said:
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.
 
WWGD said:
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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