No solutions for sqrt{x}=-5

Inertia_Squared

Inertia_Squared

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I've recently been on a blackpenredpen binge (great YouTube channel if you're interested), and in one of his videos he stated that [math]\sqrt{x} = -5[/math] has no solution for x, and even says there are no complex solutions. However if you let [math]x = 25i^4[/math] then as long as you acknowledge that [math]25 \neq 25i^4[/math] by assuming the RHS is always complex (i.e. that it cannot be further simplified, even though it obviously can) wouldn't that make it a valid solution?

[math]\text{Since } x = 25i^4, \text{then} \sqrt{x} = 5i^2 = -5[/math]
I know that this is a problem since sqrt{25} is not equal to -5, but if you treat 25 and 25i^4 as separate values I can't see how this causes any issues if you say it is a purely complex solution. I'm sure this is probably wrong, so if anyone knows a solid example that contradicts this assumption I would love to see it!
 
Just realized I've posted in the wrong section, sorry about that! If this needs to be posted elsewhere I'm happy to do so, apologies for the inconvenience!
 
[math]\sqrt {xy} \neq \sqrt x\sqrt y[/math] if x and y are both negative.

That is [math] \sqrt{i^4} = \sqrt {i^2i^2} \neq \sqrt {i^2} \sqrt {i^2}[/math]
 
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Inertia_Squared said:
Just realized I've posted in the wrong section, sorry about that! If this needs to be posted elsewhere I'm happy to do so, apologies for the inconvenience!
Click to expand...
Please don't repost this in another category. It is fine where it is and besides, I asked the SUPERmoderator to move the post.
 
Inertia_Squared said:
I've recently been on a blackpenredpen binge (great YouTube channel if you're interested), and in one of his videos he stated that [math]\sqrt{x} = -5[/math] has no solution for x, and even says there are no complex solutions. However if you let [math]x = 25i^4[/math] then as long as you acknowledge that [math]25 \neq 25i^4[/math] by assuming the RHS is always complex (i.e. that it cannot be further simplified, even though it obviously can) wouldn't that make it a valid solution?

[math]\text{Since } x = 25i^4, \text{then} \sqrt{x} = 5i^2 = -5[/math]
I know that this is a problem since sqrt{25} is not equal to -5, but if you treat 25 and 25i^4 as separate values I can't see how this causes any issues if you say it is a purely complex solution. I'm sure this is probably wrong, so if anyone knows a solid example that contradicts this assumption I would love to see it!
Click to expand...
In Elementary Calculus: An Infinitesimal Approach by Jerome Keisler there is discussion from modal theory building an enlarged number system.
This can be found HERE . There one can down load the text in PdF format free.
In the spirit of that approach there is a group of analyst who propose that we define [imath]\mathscr{i}[/imath] as a "new" number that solves the equation [imath]x^2+1=0[/imath], Note that [imath]-\mathscr{i}[/imath] also solves the same equation. Also note that there is no mention of a radical symbol, [imath]\sqrt{\;}[/imath] . We have not intention dropping the use of radical symbol. only to say that it applies to non-negative real numbers. thus, [imath]\sqrt{x^2+1}[/imath] tells the user that [imath]x^2+1[/imath] is a non-negative real number.
Now many many of us "grew up mathematically" saying that [imath]{\bf i}=\sqrt{-1}[/imath] but that leads to problems such as you noted.
In the new number system we have [imath]\mathscr{i}^2=-1,~\mathscr{i}^3=-\mathscr{i}~\&~\mathscr{i}^4=1[/imath] as usual.

Now I have no notion of giving a course on complex variables here. But know this: every complex number has n nth roots.
There are four fouth roots of [imath]-16[/imath]. SEE HERE



[imath][/imath][imath][/imath][imath][/imath][imath][/imath]
 
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