Well, About 3-5 percent of the best students in a cohort can expect to get First Class Honours. It basically means 97th percentile, or 95th percentile, depending on the quality of the students. The 75th to 95th percentile can expect to get Second Class Honours.
Which implies that I can, tentatively, estimate you to be in the top 10% of people who are accepted for a degree. That’s really good.
I must admit that this question stunned me. I don’t actually know.
...I think we’ve found the start of the problem. Your foundations have a few holes.
Dividing X by Y, at its core, means that I have X objects, I want to place them in Y exactly equal piles, how many objects do I place per pile? (At least, that’s the definition I’d use). In this way, the usefulness of the operation is immediately apparent; if I have six apples, and I want to divide them among three people, I can give each person two apples.
I can use the same definition if I have five apples and three people; then I give each person one and two-thirds apples.
This also works for negative numbers; if I have negative-six apples (i.e. a debt of six apples) I can divide that into three piles by placing negative-two apples in each pile.
Division by zero then becomes a matter of taking (say) six apples, and trying to put them into zero piles. (I hope that makes the problem with division by zero clear).
And yes, there is a fancy algorithm that I can put X and Y in and get the quotient out… but that algorithm is not a particularly good basic definition of division. (Interestingly, I note that your definition jumps straight to setting out separate cases and then trying to apply a different algorithm to each individual case. This would make it very hard to work with in practice; I’ve worked with division algorithms on computers, and they’re far simpler, conceptually, than what you had there. If that’s what you’ve been working with, then I am really not surprised that you’ve been having trouble with maths).
Now let’s see how far this goes...
Define “multiplication”, “addition”, and “subtraction”.
When I read your answer, I was thinking, (seriously no offense because I know you are really smart) I don’t know for sure that this definition works for complex numbers.
It does; complex numbers are just another type of number. We’ll get to them shortly.
And then I was thinking that mathematics relies on definitions and deductive reasoning and intuition cannot give the certainty of deductive reasoning, thus it might be a fallacy to think that something simple and intuitive is an accurate model of mathematical reality… then I remembered that it was taught in kindergartens even...
To be fair, sometimes the intuitive answer is wrong; one does have to take care. But sometimes, as in these cases, the intuitive model does work.
Define “multiplication”
X*Y : I have Y sets of X objects, how many objects do I have?
Exactly.
“addition”
X+Y : I have X objects. I am given Y objects. How many objects do I have?
Perfect.
It’s easy to visualize imaginary numbers as another type of object ‘x’, and I am given y objects. So I have x + y imaginary objects and X + Y real objects.
You could do it that way, and it leads to the correct answers, but I think it’s fundamentally problematic to see complex numbers as intrinsically different to real numbers. (For one thing, real numbers are a subset of complex numbers in any case).
“subtraction”
X-Y : I have X objects. Y objects are taken away from me.
Right.
Then it makes me wonder what other exceptions to manipulation there is
There’s only one that I can think of off the top of my head; if x^z=y^z, this does not mean that x=y (i.e. we can’t just take the z’th root on both sides of the equation). This can be clearly demonstrated with x=2, y=-2 and z=2. Two squared is four, which is equal to (negative two) squared, but two is not equal to negative two.
Now, as to complex numbers. Let me start by asking you to define a “complex number”.
Okay, those are all—well, I think I can kind of see some relation to complex numbers in there, but it’s very vague.
So, let me describe how I understand complex numbers. To do that, we’ll have to go right back to the very basics of mathematics; numbers.
Imagine, for a moment, an infinite piece of paper. (Or you can get a piece of paper and draw this, if you like; you won’t need to draw the whole, infinite thing, just enough to get the idea)
Take a point, nice and central. Mark it “zero”.
Select a second point (traditionally, this point is chosen to the right of zero, but the location doesn’t matter). Mark it “one”.
Now, let us call the distance between zero and one a “jump”. You start from zero, you move a jump in a particular direction, you get to “one”. You move another jump in the same direction, you get to “two”. Another jump, “three”. Another jump, “four”. And so on, to infinity. These are the positive integers.
Now, consider an operation; addition. If I apply addition to any pair of positive integers, I get another positive integer. Any of these numbers that I add gives me a number I already have; I can add no new numbers with addition.
However, I can also invert the addition operation, to get subtraction. If I want to find X+Y, I hop X jumps from the zero point,then Y more jumps. But if I want to find X-Y, I must jump X jumps to the right, then Y jumps to the left; and this gives me the negative integers. Add them to the mental numberline.
At this point, multiplication gives us no new numbers. Division , however, does.
You will now notice, there are still gaps between the numbers. To fill these gaps, we turn to division; X/Y gives us a plethora of new numbers (1/2, 2⁄3, 3⁄4, 4⁄5, so on and so forth), hundreds and millions and billions of little dots between each point on the numberline. These are the rational numbers.
Is the numberline full yet? Hardly; it turns out that the rational numbers are so small a proportion of the numberline that it’s still more empty space than marked point. I could say that there’s billions of irrational numbers for every rational number, but that severely underestimates the number of irrational numbers that there are.
But let’s add all the irrational numbers as well. (If you’re actually drawing this, just take a ruler and draw a line across the page, such that all your integers fall on the line).
This line, then, is the famous numberline. I’m sure you’ve seen it before, on classroom walls and similar. It contains all the real numbers and, now that we’ve added the irrational numbers, it is full; there is no space on the line where another number can be added.
Now, let’s consider squaring. The square of one is one. The square of any positive number greater than one is an even greater positive number (for example, two squared is four). The square of any positive number between zero and one is a positive number closer to zero (0.5 squared is 0.25).
The square of zero is zero.
The square of any negative number is equal to the square of the corresponding positive number; thus the square of negative two is four.
Therefore, four has two square roots; 2 and −2. Similarly, one’s square roots are one and minus one.
So, a question then emerges; where are the square roots of minus one?
They cannot be on the numberline. There is no space for new numbers on the line, and the square of every number on the line is a positive number (or zero).
Let us call the square roots of minus one i and -i (somwhat arbitrary notation that was used once and stuck) Where do we put them on the line?
Since the line is full, we cannot put them on the line. If you place the line such that the zero is in front of you, the positive numbers head off to the right, and the negative numbers go to the left, then i is found one jump directly up from zero. Similarly, -i is one jump directly down from zero.
So, they are numbers, but they are not on the number line.
And now that we have placed i and -i, we can apply the same operation as we used earlier.
Addition: adding 1 is a jump to the right. Similarly, adding i is a jump upwards. There is a 2i two jumps above zero; a 3i three jumps above zero, and so on.
In fact, by following the same steps as were used to construct the original, real number line, we can create an imaginary number line at right angles to it; so that we can point to, say, 2.5i, or even pi i.
Then, if we want t find the point where (say) 3+4i is, we first jump three jumps to the right, then we jump four jumps up; adding the numbers 3 and 4i. 3+4i is thus a clearly defined point on the numberplane (since it’s no longer one-dimensional, “numberline” is not exactly accurate anymore).
Adding and subtracting complex numbers on this plane is perfectly straightforward (though actually describing what i apples look like is beyond me). Multiplication follows the rules for multiplying additive expressions; that is, (a+b)*(c+d) = ac+ad+bc+bd. So, therefore:
Which implies that I can, tentatively, estimate you to be in the top 10% of people who are accepted for a degree. That’s really good.
...I think we’ve found the start of the problem. Your foundations have a few holes.
Dividing X by Y, at its core, means that I have X objects, I want to place them in Y exactly equal piles, how many objects do I place per pile? (At least, that’s the definition I’d use). In this way, the usefulness of the operation is immediately apparent; if I have six apples, and I want to divide them among three people, I can give each person two apples.
I can use the same definition if I have five apples and three people; then I give each person one and two-thirds apples.
This also works for negative numbers; if I have negative-six apples (i.e. a debt of six apples) I can divide that into three piles by placing negative-two apples in each pile.
Division by zero then becomes a matter of taking (say) six apples, and trying to put them into zero piles. (I hope that makes the problem with division by zero clear).
And yes, there is a fancy algorithm that I can put X and Y in and get the quotient out… but that algorithm is not a particularly good basic definition of division. (Interestingly, I note that your definition jumps straight to setting out separate cases and then trying to apply a different algorithm to each individual case. This would make it very hard to work with in practice; I’ve worked with division algorithms on computers, and they’re far simpler, conceptually, than what you had there. If that’s what you’ve been working with, then I am really not surprised that you’ve been having trouble with maths).
Now let’s see how far this goes...
Define “multiplication”, “addition”, and “subtraction”.
It does; complex numbers are just another type of number. We’ll get to them shortly.
To be fair, sometimes the intuitive answer is wrong; one does have to take care. But sometimes, as in these cases, the intuitive model does work.
Exactly.
Perfect.
You could do it that way, and it leads to the correct answers, but I think it’s fundamentally problematic to see complex numbers as intrinsically different to real numbers. (For one thing, real numbers are a subset of complex numbers in any case).
Right.
There’s only one that I can think of off the top of my head; if x^z=y^z, this does not mean that x=y (i.e. we can’t just take the z’th root on both sides of the equation). This can be clearly demonstrated with x=2, y=-2 and z=2. Two squared is four, which is equal to (negative two) squared, but two is not equal to negative two.
Now, as to complex numbers. Let me start by asking you to define a “complex number”.
Okay, those are all—well, I think I can kind of see some relation to complex numbers in there, but it’s very vague.
So, let me describe how I understand complex numbers. To do that, we’ll have to go right back to the very basics of mathematics; numbers.
Imagine, for a moment, an infinite piece of paper. (Or you can get a piece of paper and draw this, if you like; you won’t need to draw the whole, infinite thing, just enough to get the idea)
Take a point, nice and central. Mark it “zero”.
Select a second point (traditionally, this point is chosen to the right of zero, but the location doesn’t matter). Mark it “one”.
Now, let us call the distance between zero and one a “jump”. You start from zero, you move a jump in a particular direction, you get to “one”. You move another jump in the same direction, you get to “two”. Another jump, “three”. Another jump, “four”. And so on, to infinity. These are the positive integers.
Now, consider an operation; addition. If I apply addition to any pair of positive integers, I get another positive integer. Any of these numbers that I add gives me a number I already have; I can add no new numbers with addition.
However, I can also invert the addition operation, to get subtraction. If I want to find X+Y, I hop X jumps from the zero point,then Y more jumps. But if I want to find X-Y, I must jump X jumps to the right, then Y jumps to the left; and this gives me the negative integers. Add them to the mental numberline.
At this point, multiplication gives us no new numbers. Division , however, does.
You will now notice, there are still gaps between the numbers. To fill these gaps, we turn to division; X/Y gives us a plethora of new numbers (1/2, 2⁄3, 3⁄4, 4⁄5, so on and so forth), hundreds and millions and billions of little dots between each point on the numberline. These are the rational numbers.
Is the numberline full yet? Hardly; it turns out that the rational numbers are so small a proportion of the numberline that it’s still more empty space than marked point. I could say that there’s billions of irrational numbers for every rational number, but that severely underestimates the number of irrational numbers that there are.
But let’s add all the irrational numbers as well. (If you’re actually drawing this, just take a ruler and draw a line across the page, such that all your integers fall on the line).
This line, then, is the famous numberline. I’m sure you’ve seen it before, on classroom walls and similar. It contains all the real numbers and, now that we’ve added the irrational numbers, it is full; there is no space on the line where another number can be added.
Now, let’s consider squaring. The square of one is one. The square of any positive number greater than one is an even greater positive number (for example, two squared is four). The square of any positive number between zero and one is a positive number closer to zero (0.5 squared is 0.25).
The square of zero is zero.
The square of any negative number is equal to the square of the corresponding positive number; thus the square of negative two is four.
Therefore, four has two square roots; 2 and −2. Similarly, one’s square roots are one and minus one.
So, a question then emerges; where are the square roots of minus one?
They cannot be on the numberline. There is no space for new numbers on the line, and the square of every number on the line is a positive number (or zero).
Let us call the square roots of minus one i and -i (somwhat arbitrary notation that was used once and stuck) Where do we put them on the line?
Since the line is full, we cannot put them on the line. If you place the line such that the zero is in front of you, the positive numbers head off to the right, and the negative numbers go to the left, then i is found one jump directly up from zero. Similarly, -i is one jump directly down from zero.
So, they are numbers, but they are not on the number line.
And now that we have placed i and -i, we can apply the same operation as we used earlier.
Addition: adding 1 is a jump to the right. Similarly, adding i is a jump upwards. There is a 2i two jumps above zero; a 3i three jumps above zero, and so on.
In fact, by following the same steps as were used to construct the original, real number line, we can create an imaginary number line at right angles to it; so that we can point to, say, 2.5i, or even pi i.
Then, if we want t find the point where (say) 3+4i is, we first jump three jumps to the right, then we jump four jumps up; adding the numbers 3 and 4i. 3+4i is thus a clearly defined point on the numberplane (since it’s no longer one-dimensional, “numberline” is not exactly accurate anymore).
Adding and subtracting complex numbers on this plane is perfectly straightforward (though actually describing what i apples look like is beyond me). Multiplication follows the rules for multiplying additive expressions; that is, (a+b)*(c+d) = ac+ad+bc+bd. So, therefore:
(3+4i)*(2+5i) = (3*2)+(3*5i) + (4i*2) + (4i*5i) = 6 + 15i + 8i + 20*i*i
But since i is defined such that i*i=-1, that means:
(3+4i)*(2+5i) = 6 + 15i + 8i + 20(-1) = 23i-14
Voila, multiplication.