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 √-1 ≠ 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∙i, or -1 > 0. Meaning that: i ≯ 0.
Suppose that i < 0, then: i² > 0∙i -- multiplying both sides with a negative number shifts the comparison symbol --, or -1 > 0. Meaning that: i ≮ 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.
So you really did fail Algebra 201?
Too bad. Well, you tried.
Originally posted by dadudemon
So you really did fail Algebra 201?Too bad. Well, you tried.
Either way, courses with names such as calculus 101 and algebra 201 are of substandard quality. They cover less content and details over more time. It's the layman's math courses.
In general, I'm better at math than you.
Authority aside, you ignored my proofs.
Originally posted by Astner
Can't you read? I never took a course called algebra 201 and thus I couldn't have failed it.Either way, courses with names such as calculus 101 and algebra 201 are of substandard quality. They cover less content and details over more time. It's the layman's math courses.
In general, I'm better at math than you.
Authority aside, you ignored my proofs.
Originally posted by Astner
Either way, courses with names such as calculus 101 and algebra 201 are of substandard quality.
That's...my point. 😉
Originally posted by Astner
They cover less content and details over more time. It's the layman's math courses.
Exactly.
Originally posted by Astner
In general, I'm better at math than you.
I agree.
Originally posted by Astner
Authority aside, you ignored my proofs.
So you really did fail Algebra 201?
Originally posted by Dolos
Solve for76x-8y+3z^2=7
3x+3y+8z=6
y+89z=-3
6x=76y+zUse distribution as opposed to elimination. 😈
Personally, I wouldn't even want to touch this problem.
Through row reduction the three lower equations will give you the following values for x, y, and z:
x = 22637862/11664162
y = 3357/21843
z = 135827172/11664162 - 255132/21843
Put the values in the top equation and you'll get that the left hand side will equal 13.525... and not 7 like the right hand side suggests.
Originally posted by Dolos
Oops, if you're confused find X, Y, Z.The answer will by like (X,Y,Z); think a three dimensional graph.
Originally posted by Astner
That's an overdetermined system—you have more equations than variables—it's not solvable.Through row reduction the three lower equations will give you the following values for x, y, and z:
x = 22637862/11664162
y = 3357/21843
z = 135827172/11664162 - 255132/21843Put the values in the top equation and you'll get that the left hand side will equal 13.525... and not 7 like the right hand side suggests.
There are no values that satisfies the equations, thus nothing to span a three-dimensional body.
That's why they never went ever three rows in the problems I had.
For four rows I guess you'd need w, x, y, z. Four variables, and a four dimensional graph???
Maybe I'm just being pseudo-intellectual right now. But would that make a 4d graph? If you had 4 variables and 4 rows????
Originally posted by Dolos
That's why they never went ever three rows in the problems I had.For four rows I guess you'd need w, x, y, z. Four variables, and a four dimensional graph???
Maybe I'm just being pseudo-intellectual right now. But would that make a 4d graph? If you had 4 variables and 4 rows????
If you have one less unique equation than you do variables, you'll get a solution in terms of a single variable function that will manifest the answers as a curve.
If you have two less unique equations than you do variables, you'll get a solution in terms of a two-variable function that will manifest the answers as a surface.
The next step is will represent the solutions as a three-dimensional body, a four-dimensional body, etc.
Originally posted by Astner
If you have as many unique equations as you do variables you'll get one solution, this will manifest itself as a dot.If you have one less unique equation than you do variables, you'll get a solution in terms of a single variable function that will manifest the answers as a curve.
If you have two less unique equations than you do variables, you'll get a solution in terms of a two-variable function that will manifest the answers as a surface.
The next step is will represent the solutions as a three-dimensional body, a four-dimensional body, etc.
Curve as in parabola?
What do you know about the fourth dimension in terms of math? Do they ever go over that in any of your courses?
Originally posted by Dolos
Curve as in parabola?
Let's examine the following basic equation.
x + 2y = 5
Where x and y are real numbers.
Two variables, one equation. This is called an underdetermined system.
Evidently there are infinitely many solutions to the equation above each of which can be expressed as x = f(y) = 2y - 5.
This function would describe x as a function of y as a straight line, with each point of the line being a solution to the equation.
So if y = 0 then x = -5. This is a solution to the equation. If y = 0.1 then x = -4.8, and so on.
So (x, y) = (-5, 0) and (-4.8, 0.1) are two of the infinitely many solutions.
Originally posted by Dolos
What do you know about the fourth dimension in terms of math? Do they ever go over that in any of your courses?
Because the answer to both questions are yes. In fact we worked with curve- and surface integrals than bended and twisted in n dimensions in one of the first courses we took.
Originally posted by Astner
Curve as in a line that can bend in any direction.Let's examine the following basic equation.
x + 2y = 5
Where x and y are real numbers.
Two variables, one equation. This is called an underdetermined system.
Evidently there are infinitely many solutions to the equation above each of which can be expressed as x = f(y) = 2y - 5.
This function would describe x as a function of y as a straight line, with each point of the line being a solution to the equation.
So if y = 0 then x = -5. This is a solution to the equation. If y = 0.1 then x = -4.8, and so on.
So (x, y) = (-5, 0) and (-4.8, 0.1) are two of the infinitely many solutions.
Yes, we just finished up that part of College Algebra. Now we're moving onto exponents. Haven't look at the homework yet.
The fourth dimension as in the mathematically Euclidean sense (or alternative geometries), or the fourth dimension in the sense of physics as expressed in Quantum Field Theory?Because the answer to both questions are yes. In fact we worked with curve- and surface integrals than bended and twisted in n dimensions in one of the first courses we took.
Nice.
Originally posted by Dolos
Yes, we just finished up that part of College Algebra. Now we're moving onto exponents. Haven't look at the homework yet.
Originally posted by Dolos
Nice.
Originally posted by Astner
College?
Like I said it's a Junior College, but not all courses are remedial. I tested out of all my Remedial Courses, so I'm still taking college level courses. Also I should earn about 13,000 dollars this year as a Mover Helper, which will help with more courses. 😉
In Euclidean geometry the transition from three to four dimensions is the same as that from two to three dimensions, you just add another axis orthogonal to the rest of them.
Therein lies the fundamental difference between the extra-dimensional concept.
Originally posted by Dolos
Like I said it's a Junior College, but not all courses are remedial. I tested out of all my Remedial Courses, so I'm still taking college level courses. Also I should earn about 13,000 dollars this year as a Mover Helper, which will help with more courses. 😉
Originally posted by Dolos
Therein lies the fundamental difference between the extra-dimensional concept.