/sci/ - Science & Math

Yes, the answer is false.

Ok, well anyone have a real answer?

>>1717464

Yeah don't listen to him, the answer is true.

>>1717464
Absolute value is always non-negative retard.

The absolute of any number is positive. A positive number cannot be less than three. There is no real or imaginary solution, only the null set.

x - 1 < - 3
x < 2

or

- x + 1 < -3
- x < -4
x > 4

>>1717472


Modulus

>>1717455
What do you mean?

>>1717460
>>1717469
Ignore these faggots. The answer is 42.

>>1717473
I dont understand why it cant be negitive?

>>1717478
Because it's defined to never be.

It's like asking why 2 is greater than 1.

>>1717480
2 is greater then 1 because that is the way I was tought to count. What do you mean "never defined to be"

>>1717480

Stop being a dick and explain it to him.

2 is only greater then 1 if you count backwards fag

>>1717478 why?
http://en.wikipedia.org/wiki/Absolute_value
That's why.

>>1717496
Ignore this guy. Wiki said this article was edited 2 minutes ago, he changed it to make you believe.

>>1717496
>>1717496
I fixed this page so it is correct.

>>1717493
Fine

|x| is defined as the greatest of the numbers -x and x. If x is negative then -x is positive and obviously greater, so the absolute value can never be negative.

That the mathematical function explicitly defined to replace all negative numbers with corresponding positive numbers can somehow produce a negative number.

>>1717504

>absolute values are always negitive

if you're going to try troll wikipedia, at least spell right

>>1717509
Yeah, you forgot the q in "negitiqves"

>>1717507
If x=2i, which is greater? x or -x?

>>1717504
>>1717504
This asshole is dicking with wikipedia. Use http://mathworld.wolfram.com/AbsoluteValue.html

wolfram alpha is always wrong

>>1717519
A positive imaginary number is still greater than a negative imaginary number.

>>1717519
Ok, that definition is only appropriate for real numbers (and obviously subsets of them).

For more general vector spaces (yes the reals and complex numbers are vector spaces) you could define the absolute value as
|x| = sqrt(<x, x>)
which is a well defined real number since the inner product is required to be positive definite. Square roots of non-negative numbers are obviously always non-negative so so is the absolute value. For C^n the inner product <u, w> is usually taken to be u_iw^i* where w* denotes the complex conjugate of w and Einstein summation convention is used. For C^1 then we would have |z|^2 = zz*, or if z = a+bi, |z| = zz* = (a+bi)(a-bi) = a^+b^2, thus |z| = sqrt(a^2+b^2). Again, a non-negative number.

IS THAT ENOUGH ALGEBRA FOR YOU

>>1717455
What if its -|x| does that make it -ve

>>1717527
Which is greater? (3+4i) or (4+3i)

>>1717531
While everything you said here is true, just stating |z| = sqrt(a^2+b^2) or |z|^2 = zz* would have been enough of a definition. You'll never get those five minutes back now.

>>1717563
Then you would ask what the absolute value of the vector (2i+3, 5-6i) is, so I decided to be as completely overkillingly rigorous as I could.

Also >implying I type that slow

>>1717570
What is the absolute value of the vector (2i+3, 5-6i)?

>>1717586
<span class="math">|\mathbf{u}|^2 = u_1u_1^* + u_2u_2^* = (-2i+3)(2i
+3) + (5-6i)(5+6i) = 2^2 + 3^2 + 5^2 + 6^2 = 4 + 9 + 25 + 36 = 76[/spoiler]
So <span class="math">\sqrt{76}[/spoiler] if I didn't screw up my arithmetic.

Wanna go full mathematician and go for complex function spaces?

>>1717586
<span class="math">|\mathbf{u}|^2 = u_1u_1^* + u_2u_2^* = (-2i+3)(2i
+3) + (5-6i)(5+6i) = 2^2 + 3^2 + 5^2 + 6^2 = 4 + 9 + 25 + 36 = 74[/spoiler]
So <span class="math">\sqrt{74}[/spoiler] if I didn't screw up my arithmetic.

Wanna go full mathematician and go for complex function spaces?

>>1717614
f(w,z)=(w^2*exp(sin(z)), w/arctan(z))
|f|=?

>>1717634
arctan(z) is the multivalued inverse tangient btw; be sure to give all solutions

>>1717644
>>1717634
Ah but you need to define the inner product we're going to use, too.

I am trying to find the general sol. of the second order nonhomogenous diff eq.

y" +8y' +16y = 8x^2 +4x +1
So far I have done this, but I am not sure if I am on the right track...

y" +8y' +16y = 8x^2 +4x +1
= r^2 +8r +16

complete square so that r = -4

I assume I can use the identity of y = e^rx (A+Bx)
= e^(-4x) . (A+Bx)

Where do I go from there?

>>1717761
Your RHS is a second degree polynomial, so attack it with a polynomial.

>>1717761
>thread starts with a basic as fuck question
>gets to this
/sci/, I am impressed

>>1717777
>implying linear DEs aren't basic as fuck

>>1717778
compared to |x-1|<-3