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I ran a search, but, oddly enough, I can't to find a similar question on here. (If so, kindly point me in that direction, and I'll take this one down.) It seems like a pretty basic question in real analysis, but I'm struggling to come up with a suitable proof that no real number satisfies $x^{2} = -1$. I assume it's a proof by contradiction, but I'm just not seeing it.

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    $\begingroup$ Show that $x^2 \geq 0$ for all real $x$. Look at 2 different cases. $\endgroup$ – Listing Feb 13 '14 at 20:44
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    $\begingroup$ Depending on how deep you want to take this, you can read about ordered fields. I mean, if you are willing to accept that the product of two positive numbers is positive, and of two negative numbers is positive, then that is enough. If you want to understand more about why those statements are true for real numbers, that link may help. $\endgroup$ – alex.jordan Feb 13 '14 at 20:49
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    $\begingroup$ if there is, you are imagining it. :) $\endgroup$ – Lost1 Feb 13 '14 at 22:22
  • $\begingroup$ Can also be proved directly with a little bit of complex algebra. See below $\endgroup$ – user76568 Feb 16 '14 at 3:09
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Assume there actually is a real number satisfying $x^2=-1$. Since $0^2=0$, it follows that $x \neq 0$. Assume $x > 0$. From $$x \cdot x= x^2 =-1$$ We obtain (Dividing by $x$): $$0<x=-\frac{1}{x}<0$$ Contradiction.
Assume $x < 0$. Again we obtain: $$0>x=-\frac{1}{x}>0$$
Contradiction.

The conclusion (The only thing that would make the above implication be true as a whole) is that there does not exist a real number $x$ for which $x^2=-1$.

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  • $\begingroup$ @user96966 I was wondering: Is this clear to you? If not, I'd be happy to be more thorough in my answer (In math, one can always be more thorough) $\endgroup$ – user76568 Feb 13 '14 at 21:25
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    $\begingroup$ Would the down voter (there is one) please explain what is wrong with the answer? So I can fix it? $\endgroup$ – user76568 Feb 13 '14 at 22:32
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    $\begingroup$ Nice! For some reason, looking at this really drives home to me how deliciously weird $i$ is. It is equal to its own negative reciprocal. Add to that the fact that $\mathbb C$ could have been defined with $-i$ and $i$ completely swapped with each other and it would have no difference from regular $\mathbb C$, and it practically makes me light-headed to think about it too much. $\endgroup$ – Pat Muchmore Feb 13 '14 at 22:39
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    $\begingroup$ @PatMuchmore $-i=i^{-1}$ is used a lot as a trick in order to simplify expressions. I love it as well :-) $\endgroup$ – user76568 Feb 13 '14 at 22:43
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Can you prove:

Proposition $(-1)(-1)=1$

Proposition If $\ a<0\ $ then $\ a=(-1)|a|$

Assuming these, if $a\ge 0$ then it is clear $a^2\ge 0$. If $a<0$ then

$$a^2=a\cdot a = (-1)|a|\cdot (-1)|a| = (-1)(-1)|a|^2 = |a|^2 > 0$$

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Here's a proof that uses calculus.

We show the function $f(x) = x^2 + 1$ has no real roots. Its derivative is $f'(x) = 2x$, so its only critical point occurs when $x=0$. Since $f''(x) = 2$, by the second derivative test, $x=0$ is a global minimum. Since $f(0) = 1$, for all $x\in\mathbb{R}$, $f(x)\geq 1 > 0$, so no real number satisfies $f(x) = 0$.

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If possible suppose that there is a non zero real number, say $r$, with $r^2=-1$

Now if $r>0$ then $-1=r^2>0$ (since multiplying both side by $r$ does not change the inequality) which is a contradiction, and if $r<0$ then multiplying by a negative number reverse the inequality and as in the first case you will get a contradiction.

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If $x>0$, then $x\cdot x>0$ since the product of two positive numbers is positive.

If $x=0$, $\;x\cdot x=0\cdot0=0\cdot0+(0\cdot0-(0\cdot0))=(0+0)\cdot0-(0\cdot0)=0\cdot0-(0\cdot0)=0$.

If $x<0$, then $-x>0$ and \begin{align} &(-x)(-x)\\=&(-x)(-x)+0\\=&(-x)(-x)+(0\cdot x-(0\cdot x)) \\=&(-x)(-x)+((0+0)x-(0\cdot x))\\=&(-x)(-x)+((0\cdot x+0\cdot x-(0\cdot x)) \\=&(-x)(-x)+0\cdot x \\=&(-x)(-x)+(-x+x)\cdot x \\=&(-x)(-x)+(-x)\cdot x+x\cdot x \\=&(-x)(-x+x)+x\cdot x \\=&(-x)\cdot0+x\cdot x \\=&(-x)\cdot0+((-x)\cdot0-((-x)\cdot0))+x\cdot x \\=&(-x)\cdot(0+0)-((-x)\cdot0)+x\cdot x \\=&((-x)\cdot0-((-x)\cdot0))+x\cdot x \\=&x\cdot x>0 \end{align} since $(-x)(-x)>0$.

Finally, $1=1\cdot1>0$ and so, $-1<0$ which implies $-1\not\geq0$ by trichotomy.

Hopefully, the axioms that I've used are clear enough.

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Direct proof:

Assume $z \in \mathbb{C}$ obeys $z^2=-1$. So: $$z(z-\overline{z})=z^2-z \overline{z}=-1-|z|^2 \neq 0 \implies z-\overline{z} \neq 0 \iff \Im{(z)} = \frac{z-\overline{z}}{2i} \neq 0 \iff z \notin \mathbb{R}$$

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This might be cheating a bit, but you could also give reference to the Fundamental theorem of Algebra: Any $n$ degree polynomial in $\mathbb{C}[x]$ has exactly $n$ roots counted with multiplicity. If you know that $i$ and $-i$ are the two roots of $x^2 +1 = 0$ and you know that $i$ and $-i$ are not real, then there are no real roots.

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  • $\begingroup$ Why the downvote? $\endgroup$ – Thomas Aug 12 '15 at 17:17

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