/wsr/ - Worksafe Requests » Thread #617565

[15 / 3 / ?]

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Anonymous Thu 17 Jan 2019 12:56:26 No.617565 View View Reply Original Report

Quoted By: >>617583 >>617601

i get that x > +-sqrt(-3). since +-sqrt(-3) is not real number, what does that mean? what is the solution?

Anonymous

Anonymous Thu 17 Jan 2019 13:39:22 No.617573 Report

Quoted By: >>617611

It means bs.
As you say it isn't a real number, thus you cannot compare it.
Yes, it sounds weird but this is what you have when you come to a complex number (i.e. the number system that square root of minus number has its meaning.)

Anonymous

Anonymous Thu 17 Jan 2019 14:17:13 No.617583 Report

Quoted By:

>>617565

That you enter the complex numbers, sqrt(3) * i

Anonymous

Anonymous Thu 17 Jan 2019 15:07:56 No.617601 Report

Quoted By: >>617608 >>617618

>>617565
>I get
Did you by any chance square root both sides of the equation without checking validity of it? If it were x^2>-3 it would simply be x being any real number.

Anonymous

Anonymous Thu 17 Jan 2019 15:22:53 No.617608 Report

Quoted By:

>>617601
no i wasn't sure about the solution so i put the problem into symbolab.com and it gave me this shit

Anonymous

Anonymous Thu 17 Jan 2019 15:27:41 No.617611 Report

Quoted By: >>617623 >>617641

>>617573
Every number is a sum of a real number and an imaginary number, so obviously every number whose real component is equal to zero will be less than a number whose real component is greater than zero.

Anonymous

Anonymous Thu 17 Jan 2019 15:41:18 No.617618 Report

Quoted By: >>617640

>>617601
Btw, the original inequality is 5 - x^2 < 8. i did it this way:
>5 - x^2 < 8
>5 < 8 + x^2
>x^2 > 5 - 8
>x^2 > -3
>x > +-sqrt(-3)
which is complex number. the problem you see is that i don't understand why is the solution any number from R.

Anonymous

Anonymous Thu 17 Jan 2019 15:43:40 No.617621 Report

Quoted By:

ahh i get it forget about it

Anonymous

Anonymous Thu 17 Jan 2019 15:47:11 No.617623 Report

Quoted By: >>617641 >>617642 >>617648

>>617611
0+i < 1+i
i<1+i
multiply both side by i
i*i < 1*i + i*i
bc i*i = i^2 = -1, we have
-1<i+-1
0<i
multiply both side by i again
0*i < i*i
bc i*i = i^2 = -1, thus
0 < -1

Sorrt guys, it seems like I just destroyed the univers' law and we're going to die now.

Anonymous

Anonymous Thu 17 Jan 2019 16:24:16 No.617640 Report

Quoted By:

>>617618
You shouldn't have done the last step. Stay at x^2>-3. Any real number squared is greater than a negative number. If you want you can look at it as a polynominal x^2 +3 >0 .

Anonymous

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Anonymous Thu 17 Jan 2019 16:27:18 No.617641 Report

Quoted By:

>>617611
>>617623

Anonymous

Anonymous Thu 17 Jan 2019 16:30:07 No.617642 Report

Quoted By: >>617647

>>617623
Why not treat multiplying by i the same way as by negative real number and reversing the order. To be honest though is there any commonly used way of comparing complex numbers?

Anonymous

Anonymous Thu 17 Jan 2019 17:01:30 No.617647 Report

Quoted By: >>617650

>>617642
Complex number are not an ordered field.

Anonymous

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Anonymous Thu 17 Jan 2019 17:07:45 No.617648 Report

Quoted By: >>617650

>>617623
>multiply both side by i
>i*i < 1*i + i*i
You can't multiply both sides of an inequality, only an equality:

0<1
-1*0<-1*1
0<-1

Anonymous

Anonymous Thu 17 Jan 2019 17:19:25 No.617650 Report

Quoted By:

>>617648
You can multiply by a positive number. You can also multiply by a negative number if you flip the inequality sign. All of this doesn't apply to complex numbers though, because >>617647.

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