Would you give this to a girl/guy?

Love you too, sweetums.

What if they came up with "3u > i"?

Haha, nice one ^

Though, maybe, it would be sweet.
But what if they don't get it. Lol.

See : subtraction

... i swore u could only divide like terms but...w/e

ITT: Hipsters

I'm pretty sure i < 3u is wrong. It should be i does not equal 3u. It would be true if 3u > 2u though. Or if this were a sub question i = 3u would go into a second equation.

I CALL THIS MATH INTO QUESTION!

ITT: Hipsters

Oh, you.

If someone doesn't get this, I probably wouldn't be dating them anyway.

i <3 this.

I'm pretty sure i < 3u is wrong. It should be i does not equal 3u. It would be true if 3u > 2u though. Or if this were a sub question i = 3u would go into a second equation.

I CALL THIS MATH INTO QUESTION!

There's nothing wrong with it.

I'm pretty sure i < 3u is wrong. It should be i does not equal 3u. It would be true if 3u > 2u though. Or if this were a sub question i = 3u would go into a second equation.

I CALL THIS MATH INTO QUESTION!

Orly?

Son, you just got out-mathed by a computer. Not number-crunching math. Equation-solving math. How does that make you feel?

Yes, but only solves for Real numbers. I is not counted as a real number in this equation. Only u.

One other thing I like to mention at some point (doesn’t have to be now) is that there are no inequalities with imaginary numbers. You cannot meaningfully say that 1>i or 1<i. Because they cannot be graphed on a number line, they don’t really have “sizes”—they can be equal or not, but they cannot be greater than or less than each other.

http://cnx.org/content/m19423/latest/

If i was counted a real number in which has the same value of u such as b, I would gladly accept the answer. There were two ways of looking at it and I didn't want to add another "d'awww that's so cute I should give it to my sister/friend" comment to this math love thread. I wanted to ruin all your gay gay fun with straight math. Mission accomplished.

PS:Thanks for the link. Might come in handy. Could use some less demeaning tone in your voice next time Hazzy.

Yes, but only solves for Real numbers. I is not counted as a real number in this equation. Only u.



http://cnx.org/content/m19423/latest/

It's entirely possible that i here is not the square root of negative 1 but just some random variable. I guess a capital "I" would have prevented confusion.

Yes, but only solves for Real numbers. I is not counted as a real number in this equation. Only u.



http://cnx.org/content/m19423/latest/

If i was counted a real number in which has the same value of u such as b, I would gladly accept the answer. And I don't give a pomegranate that you threw a equation into a online computer and came up with the answer it told you. You did not prove your theory in any way. So how's that make you feel? Like a child being talked down to? Ya, I thought so.

PS:Thanks for the link. Might come in handy.

What if we say a+bi > c+di if |a+bi| > |c+di|? Where |a+bi| is the magnitude.

I like how everyones calling it nerdy even though i learned how to do this kind of math in 7th grade. Add some calc, then perhaps

I like how everyones calling it nerdy even though i learned how to do this kind of math in 7th grade. Add some calc, then perhaps

You'd be surprised at how many people suck at algebra. Really, it's crazy.

Yes, but only solves for Real numbers. I is not counted as a real number in this equation. Only u.

"solve for i" tells you that i is a variable, and not a constant.

PS:Thanks for the link. Might come in handy. Could use some less demeaning tone in your voice next time Hazzy.

Funhaus. Don't take anything offending from here seriously.

You'd be surprised at how many people suck at algebra. Really, it's crazy.

I suck at algebra.

What if we say a+bi > c+di if |a+bi| > |c+di|? Where |a+bi| is the magnitude.

This is why I should just shut my mouth when it comes to forums and math. No matter what answer you give there's always another question. I'll take a stab at it but I can tell you are trying to turn it into a real number by using an absolute value.

|z| = square root of (9x-7i)(9x-7i)
|z| = square root of 81x2-63xi-63xi+49
|z| = square root of 81x2-126xi+49

|z| = square root of (9x-21u)(9x-21u)
|z| = square root of 81x2-189xu-189xu+441
|z| = square root of 81x2-378xu+441
|z| = square root of 3/27x2-126xi+147

That's as far as I've gotten. I haven't done complex numbers in forever. I'll be sure to work on it and I'd appreciate it if you don't say the answer so I can keep working on it. I'm sure I'm on the wrong track anyways.

Funhaus. Don't take anything offending from here seriously.

I won't. But I truly have no idea why I even post on here. I haven't had fun in the funhouse in a long ass time. I believe I shall ban myself from it actually. Uni is starting again in 11 days and I don't have any reason to even view this forum >.>

Yes, but only solves for Real numbers. I is not counted as a real number in this equation. Only u.

Whereas right that a "i < 3u"-definition does not (given u is a real number), as far as we currently know, make any sense, does not mean that this does not make sense.

Say we define i = sqrt(-1), and say that u = - sqrt(-1), then if we substitute in:

sqrt(-1) < 3 * -(sqrt(-1))

Dividing by sqrt(-1) gives us "1 < -3", which is true.

Basically, any imaginary number lower than 1/3 * i = u satisfies this equation.

Noah

Whereas right that a "i < 3u"-definition does not (given u is a real number), as far as we currently know, make any sense, does not mean that this does not make sense.

Say we define i = sqrt(-1), and say that u = - sqrt(-1), then if we substitute in:

sqrt(-1) < 3 * -(sqrt(-1))

Dividing by sqrt(-1) gives us "1 < -3", which is true.

Basically, any imaginary number lower than 1/3 * i = u satisfies this equation.

Noah

Do you ever have fun, Noah? :f6: