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Let $i = \sqrt{-1}$ and $x \in \mathbb{R}$, then we know that $x^2\geqslant0$ for all $x \in\mathbb R$. Therefore, $$x^2 > -1$$ which implies $x > i$? Hence, every real number is greater than $i$ since $x∈ \mathbb R$. Now, this implies, $i<0$. Hence, $i$ is negative.

Now, since $i$ is negative, let $i = -m$ where $m$ is the magnitude of $i$.

We know that $i = \sqrt{-1}$. Substituting the value of $m$, we get: $-m = \sqrt{-1}$.

Squaring both the sides, we get: $$m^2= -1 \implies m = \sqrt{-1}$$ Hence, $$i = -m = -i$$ $$⇒i = -i ⇒i = 0$$

This is a contradiction to (i) that $i<0$ as at the same time, $i=0$

Then, what is $i$, or what is $ \sqrt{-1}$ exactly? What does this contradiction prove?

Angelo
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Yatharth Shrivastava
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    Why would $x^2>-1$ imply $x>\mathrm{i}$? You used a propery of $\sqrt{;}$ that holds only for non-negative real numbers. – Gary Mar 17 '23 at 07:01
  • In many cases, the source of confusion/misunderstanding is bad or ambiguous notations: $\sqrt{-1}$ is such a notation. You will not see it in any good mathematics textbook ; more precisely, it has been banned from common usage many years ago (still used sometimes by our physicists colleagues...) because bringing much confusion... – Jean Marie Mar 17 '23 at 07:04
  • On taking square root of $x² > -1$ on both sides, we get $x > √(-1)$ and hence, $x > i$ since we assumed that $i = √(-1)$. I think we do not have a problem of $±$ as we aren't solving the radical. @Gary – Yatharth Shrivastava Mar 17 '23 at 07:05
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    hey this is cool that you are experimenting with numbers. There is a first mistake in your development.

    $x^2 > -1$ does not imply $x > i$. Remember that for any $x^2 > a > 0$ you have that $x > a$ or $x < -a$. Secondly, i would say that $i < 0$ makes no sense, this is a comparison between a real number and a pure imaginary number.

     – lulufofo Mar 17 '23 at 07:06
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    @FHg I did not say that $x>-1$, I said $x>i$ and proved it in previous comment. – Yatharth Shrivastava Mar 17 '23 at 07:07
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    @YatharthShrivastava $a>b$ implies $\sqrt{a}>\sqrt{b}$ iff $b\ge 0$. In your case $b=-1$ is not non-negative. – Gary Mar 17 '23 at 07:12
  • @FHg I took it logical for your second argument. I have used the identity that $x² > -1$ to take it to $i<0$ in my question. – Yatharth Shrivastava Mar 17 '23 at 07:13
  • @Gary but why? I mean where do I get the proof of this? Any logical proof? – Yatharth Shrivastava Mar 17 '23 at 07:15
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    The square root is initially defined for non-negative numbers. If we agree that $\sqrt{x}>0$ for $x>0$, then this is a well known property. Now, when you extend the square root to negative numbers, this property will no longer be valid. Note that the complex numbers cannot even be made into an ordered field. – Gary Mar 17 '23 at 07:24
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    The problem comes from this comparison $ i < 0$ that it is not really defined as you used it in $\mathbb{C}$, look at this topic: https://math.stackexchange.com/questions/1032257/ordering-of-the-complex-numbers. – lulufofo Mar 17 '23 at 07:27
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    Thank you all. You explained my false assumptions. Once again, thank you. – Yatharth Shrivastava Mar 17 '23 at 07:30
  • I tagged the question as #recreational-mathematics earlier. There was no role of research effort in this. @theonewhodownvoted – Yatharth Shrivastava Mar 17 '23 at 07:32
  • https://math.stackexchange.com/help/someone-answers @YatharthShrivastava – Anne Bauval Mar 17 '23 at 08:18
  • @YatharthShrivastava, $x^2>-1$ just implies $x^2>i^2$ and from it you cannot deduce $x>i$. So you are wrong when you write that every real number is greater than $i$. – Angelo Sep 12 '23 at 06:50

1 Answers1

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To summarize, there are many problematic assumptions and missteps in reasoning:

  1. As pointed out in the comments, $x^2>-1$ does not imply $x>i$. The square root function is increasing over non-negative reals. Also, statements saying one complex number is greater than another (when either of them is not real) are meaningless.
    NB: As also pointed out in the comments, the complex numbers cannot be made into an ordered field under usual addition and multiplication. See here if interested in further discussion on ordering complex numbers.

  2. If for $x,a \in \mathbb{R}$, $x^2\geq a\geq 0$, we would conclude either $x\geq \sqrt{a}$ or $x\leq -\sqrt{a}.$

  3. If you did happen to show that every real number is greater than "something," that "something" would certainly not be a finite negative quantity (consider the real number that is one less than that quantity for a contradiction).

  4. Independent of the above issues, what ultimately gives the fallacious result $i=0$ is this misstep: $$-m=\sqrt{-1}\implies m^2=-1\implies m=\sqrt{-1}.$$ Instead, if $m^2=a$, we would conclude either $m=\sqrt{a}$ or $m=-\sqrt{a}$ .

Golden_Ratio
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