i = -i ?

[jrarick]

I was wondering you opinions on this proof. I know it sounds weird, but the logic seems to flow. Thanks!

​

 

[Otis]

the logic seems to flow

Hi jrarick. It's not a proof. I'm okay with the first three lines. At Line3, we have an equation that says √(-1) equals √(-1). In other words, it says i=i.

But then, the flow goes into a hole. We can see that Line4 is not a valid proportion because it violates this property: The product of the means equals the product of the extremes.

[imath]\text{Given a valid proportion } \color{blue}\frac{a}{b}=\frac{c}{d}\color{black} \quad\text{ then } \color{blue}ad=bc[/imath]

Applying this property shows a falsity: [imath]\color{red}i^2=1\color{black}[/imath].
[imath]\;[/imath]

 

[Dr.Peterson]

I was wondering you opinions on this proof. I know it sounds weird, but the logic seems to flow. Thanks!

View attachment 38333​

The problem is that every (non-zero) complex number has two square roots; and there is no definition of a primary square root by which the property you assumed in the fourth line, that [imath]\displaystyle\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}[/imath], is true.

 

[blamocur]

I was wondering you opinions on this proof. I know it sounds weird, but the logic seems to flow. Thanks!

View attachment 38333​

Pretty cool puzzle! But: once you apply non-integer powers you have to be careful about multiple values, which in the case of square roots is about signs. The third line should properly be written as [math]\sqrt{\frac{-1}{1}} = \pm \sqrt{\frac{1}{-1}}[/math]

 

[Cubist]

I was wondering you opinions on this proof. I know it sounds weird, but the logic seems to flow. Thanks!

View attachment 38333​

A rule to remember is:- you can only apply [imath]\displaystyle\sqrt{\frac{a}{b}} \to \frac{\sqrt{a}}{\sqrt{b}}[/imath] if b>0

This is related to the "power of power" reduction which can only be applied under certain circumstances. See this post (and the one after). When this reduction is initially taught pupils aren't yet ready to learn the full set of conditions when it can be applied But, for some reason, this doesn't ever seem to be revisited later during complex number study

In the OP on line 3 right hand side you have:-
[math]\sqrt{\frac{1}{-1}}[/math]And...
[math]\sqrt{\frac{1}{-1}} = (a^b)^c = ((-1)^{-1})^{(1/2)}[/math]And this does not equal...
[math] (-1)^{\left(-1 \times (1/2)\right)} = \left((-1)^{1/2}\right)^{-1} = \frac{1}{\sqrt{-1}}[/math]...because "c" on the first line is not an integer.