Moderately Difficult Math Problem (solve if you're smart enough).

Started by Barker4 pages
Barker Thorin Fan Club President

ef1

silver_tears Senior Member

Originally posted by dadudemon
parenthesis

You're close but not quite right.

i/3 < u is the correct answer but there is another way to express that same exact inequality that is the answer I am looking for.

Jaden, you're way off.

You're so lame....haermm

Astner The Ghost Who Walks

Re: Moderately Difficult Math Problem (solve if you're smart enough).

Originally posted by dadudemon
9x - 7i > 3(3x - 7u)

i = square root of -1

There is a surprise for the first person that solves the problem.


First and foremost, the imaginary unit i isn't the square root of -1. As the space of real numbers is a subspace to complex numbers, still making the square root of negative numbers nonexistent. Also, see proof.

Secondly you can't compare two complex numbers. As such a comparison wouldn't have the basic properties required of the subspace such as symmetry and reflexivity, and would thus be nonsensical.

Thoren Static Stance

You tell'em!!!

2pac4life Junior Member

Oh damn, if it has a solution, it's gonna be tough to figure it out

2pac4life Junior Member

Re: Moderately Difficult Math Problem (solve if you're smart enough).

Originally posted by dadudemon
9x - 7i > 3(3x - 7u)

i = square root of -1

There is a surprise for the first person that solves the problem.


Whats the surprise though 😮

silver_tears Senior Member

i <3 u

dadudemon Senior Member
Originally posted by silver_tears
You're so lame....haermm
Originally posted by silver_tears
i <3 u

😆 👆

You won the prize!

I love you, Irene! hug

Paranthesis, you were so very close to getting the prize, man. SO CLOSE! 😆

Astner The Ghost Who Walks

First time I've seen someone winning by being conventionally wrong.

If i wasn't the imaginary unit, and you'd reformed the relation then this could've worked out.

dadudemon Senior Member

Originally posted by Astner
First and foremost, the imaginary unit i isn't the square root of -1. As the space of real numbers is a subspace to complex numbers, still making the square root of negative numbers nonexistent. Also, see proof.

Secondly you can't compare two complex numbers. As such a comparison wouldn't have the basic properties required of the subspace such as symmetry and reflexivity, and would thus be nonsensical.

Originally posted by Astner
First time I've seen someone winning by being conventionally wrong.

If i wasn't the imaginary unit, and you'd reformed the relation then this could've worked out.

Astner, I had no idea you failed College Algebra 201. 🙁

parenthesis Restricted

Originally posted by Astner
First time I've seen someone winning by being conventionally wrong.

If i wasn't the imaginary unit, and you'd reformed the relation then this could've worked out.

I agree, the additional statement about i made me first think it said "it wasn't the root of minus 1", then when I read it again I thought the problem was to find the value of u as you would any algebraic problem. To that, I only have one thing left to say.

Originally posted by Barker
ef1
silver_tears Senior Member

Originally posted by dadudemon
😆 👆

You won the prize!

I love you, Irene! hug

blush2

Ax3l Cummin Soon

Originally posted by Astner
First time I've seen someone winning by being conventionally wrong.
You're not an American voter, are you.

Thoren Static Stance

he's like from estonia or something.

Barker Thorin Fan Club President

i'm going to drop toe hold all of you

Astner The Ghost Who Walks

Originally posted by dadudemon
Astner, I had no idea you failed College Algebra 201. 🙁

College algebra? We dealt with complex numbers in high school, and even then I was taught that what you ask is wrong.

That said, I've passed all math classes I've had, ranging from single variable analysis to partial differential equations - second course.

I've already proved why &#8730;-1 &#8800; i. But since you insist on an outright mathematical brawl, it's on!

So let me explain why you can't compare two complex numbers, and I'm not going to write this in TeX.

Suppose that i > 0, then: i² > 0&#8729;i, or -1 > 0. Meaning that: i &#8815; 0.

Suppose that i < 0, then: i² > 0&#8729;i -- multiplying both sides with a negative number shifts the comparison symbol --, or -1 > 0. Meaning that: i &#8814; 0.

In other words complex numbers can only be equal to each other, not compared, as the imaginary unit i is an element that can't be compared.

Q.E.D. mother****er.

Bardock42 Junior Member

Stalky the Clown answered this on my forum right after you posted it, and, parenthesis actually answered it as well:

Originally posted by parenthesis
-7i > -21u

-i > -3u

i < 3u

-1 < 9u(sq)

1 > -9u(sq)

1/9 > -u(sq)

1/3 > -u

-1/3 < u

I think.

Bardock42 Junior Member
Originally posted by Ax3l
You're not an American voter, are you.

Ohohoho, classic political comedy, epic!

Bardock42 Junior Member

Originally posted by Astner
College algebra? We dealt with complex numbers in high school, and even then I was taught that what you ask is wrong.

That said, I've passed all math classes I've had, ranging from single variable analysis to partial differential equations - second course.

I've already proved why &#8730;-1 &#8800; i. But since you insist on an outright mathematical brawl, it's on!

So let me explain why you can't compare two complex numbers, and I'm not going to write this in TeX.

Suppose that i > 0, then: i² > 0&#8729;i, or -1 > 0. Meaning that: i &#8815; 0.

Suppose that i < 0, then: i² > 0&#8729;i -- multiplying both sides with a negative number shifts the comparison symbol --, or -1 > 0. Meaning that: i &#8814; 0.

In other words complex numbers can only be equal to each other, not compared, as the imaginary unit i is an element that can't be compared.

Q.E.D. mother****er.

Quod Erat Demonstrandum Matrem Fututor indeed

Bardock42 Junior Member

Quadruple post FTW