This doesn’t work if you have to deal with multiplication of numbers that are not integers. You can adjust your idea to work with rational numbers (i.e. ratios of integers) but you will have trouble once you start wanting to multiply irrational numbers like e and pi where you cannot treat multiplication easily as repeated addition.
The actual answer here is that the set of real numbers form a structure called an ordered field and that the nice properties we are familiar with from algebra (for ex that a product of two negatives is positive) can be proved from properties of ordered fields.
Before i ask my question, know that my math is all the way in the back of my head and i didnt get too far in math at school.
Wdym irrational numbers dont work? -3 * -pi would be the same as 3*pi, no?
I always assumed if all factors of the multiplication are negative, it results in the same as the positive variant, no matter the numbers ( real, fractal, irrational, … )
Multiplying two negative irrational numbers together will still give you a positive number, it’s just that you can’t prove this by treating multiplication as repeated addition like you can multiplication involving integers (note that 3 is an integer, 3 is not irrational, the issue is when you have two irrationals).
So, for example with e * pi, pi isn’t an integer. No matter how many times we add e to itself we’ll never get e * pi.
Try it yourself: Assume that we can add e to itself k (a nonnegative integer) times to get the value e * pi. Then e * pi = ke follows by basic properties of algebra. If we divide both sides of this equation by e we find that pi=k. But we know k is an integer, and pi is not an integer. So, we have reached a contradiction and this means our original assumption must be false. e * pi can’t be equal to e added to itself k times (no matter which nonnegative integer k that we pick).
Thanks for that! Helped understand it a bit more, i think. So its a case of it not working on irrational numbers, its just that we cant prove it because we cant calculate the multiplication of 2, right?
Somehow, my mind has issues with the e*pi = ke. Id say that ke = e * pi is impossible because k is an integer and pi isnt, no? It could never be equals, i think.
So its a case of it not working on irrational numbers, its just that we cant prove it because we cant calculate the multiplication of 2, right?
The issue is the proving part. We can’t use repeated addition trickery (at least not in an obvious way) to show a product of two irrational negative numbers is positive. It’s definitely still true that a product of two negative numbers is positive, just that proving it in general requires a different approach.
Somehow, my mind has issues with the e*pi = ke. Id say that ke = e * pi is impossible because k is an integer and pi isnt, no? It could never be equals, i think.
Yes this is correct. The ke example is for a proof by contradiction. We are assuming something is true in order to show it forces us to be able to conclude something ridiculous/false. Since the rest of our reasoning was correct, then it must have been our starting assumption that was wrong. So, we have to conclude our starting assumption was wrong/false.
Fun fact… a formal definition of irrational numbers didn’t exist until the 1880s (150+ years after Newton died). There were lots of theories before that time (including that they didn’t exist) and they were mostly ignored. Iirc, it was Euler’s formal definition of complex numbers and e (an irrational number) that led to renewed interest.
There are some subtleties to this particular topic that are worth mentioning. I would be careful to distinguish between constructing vs defining here.
The usual definition of the irrationals works roughly like this:
You have a set of numbers R which you call the real numbers. You have a subset of the real numbers Q which you call the rational numbers. You define a real number to be irrational if it is not a rational number.
This is perfectly rigorous, but it relies on knowing what you mean by R and Q.
Both R and Q can be defined “without” (a full) construction by letting R be any complete ordered field. Such a field has a multiplicative identity 1 by definition. So, take 0 along with all sums of the form 1, 1+1, 1+1+1 and so on. We can call this set N. We can take Z to be the set of all elements of N and all additive inverses of elements of N. Finally take Q to be the set containing all elements of Z and all multiplicative inverses of (nonzero) elements of Z. Now we have our R and Q. Also, each step of the above follows from our field axioms. Defining irrationals is straightforward from this.
So, the definition bit here is not a problem. The bigger issue is that this definition doesn’t tell us that a complete ordered field exists. We can define things that don’t exist, like purple flying pigs and so on.
What the dedekind cut construction shows is that using only the axioms of zfc we can construct at least one complete ordered field.
Copy pasted from my other comment:
This doesn’t work if you have to deal with multiplication of numbers that are not integers. You can adjust your idea to work with rational numbers (i.e. ratios of integers) but you will have trouble once you start wanting to multiply irrational numbers like e and pi where you cannot treat multiplication easily as repeated addition.
The actual answer here is that the set of real numbers form a structure called an ordered field and that the nice properties we are familiar with from algebra (for ex that a product of two negatives is positive) can be proved from properties of ordered fields.
Before i ask my question, know that my math is all the way in the back of my head and i didnt get too far in math at school.
Wdym irrational numbers dont work? -3 * -pi would be the same as 3*pi, no?
I always assumed if all factors of the multiplication are negative, it results in the same as the positive variant, no matter the numbers ( real, fractal, irrational, … )
Multiplying two negative irrational numbers together will still give you a positive number, it’s just that you can’t prove this by treating multiplication as repeated addition like you can multiplication involving integers (note that 3 is an integer, 3 is not irrational, the issue is when you have two irrationals).
So, for example with e * pi, pi isn’t an integer. No matter how many times we add e to itself we’ll never get e * pi.
Try it yourself: Assume that we can add e to itself k (a nonnegative integer) times to get the value e * pi. Then e * pi = ke follows by basic properties of algebra. If we divide both sides of this equation by e we find that pi=k. But we know k is an integer, and pi is not an integer. So, we have reached a contradiction and this means our original assumption must be false. e * pi can’t be equal to e added to itself k times (no matter which nonnegative integer k that we pick).
Thanks for that! Helped understand it a bit more, i think. So its a case of it not working on irrational numbers, its just that we cant prove it because we cant calculate the multiplication of 2, right? Somehow, my mind has issues with the e*pi = ke. Id say that ke = e * pi is impossible because k is an integer and pi isnt, no? It could never be equals, i think.
The issue is the proving part. We can’t use repeated addition trickery (at least not in an obvious way) to show a product of two irrational negative numbers is positive. It’s definitely still true that a product of two negative numbers is positive, just that proving it in general requires a different approach.
Yes this is correct. The ke example is for a proof by contradiction. We are assuming something is true in order to show it forces us to be able to conclude something ridiculous/false. Since the rest of our reasoning was correct, then it must have been our starting assumption that was wrong. So, we have to conclude our starting assumption was wrong/false.
3 pi = pi + pi +pi
Sure thats okay, but what about e * pi?
I think all they mean is you can’t write it out since irrational numbers have no end.
You’re correct in that the principle still applies in exactly the same way.
Fun fact… a formal definition of irrational numbers didn’t exist until the 1880s (150+ years after Newton died). There were lots of theories before that time (including that they didn’t exist) and they were mostly ignored. Iirc, it was Euler’s formal definition of complex numbers and e (an irrational number) that led to renewed interest.
E: Richard Dedekind
There are some subtleties to this particular topic that are worth mentioning. I would be careful to distinguish between constructing vs defining here.
The usual definition of the irrationals works roughly like this:
You have a set of numbers R which you call the real numbers. You have a subset of the real numbers Q which you call the rational numbers. You define a real number to be irrational if it is not a rational number.
This is perfectly rigorous, but it relies on knowing what you mean by R and Q.
Both R and Q can be defined “without” (a full) construction by letting R be any complete ordered field. Such a field has a multiplicative identity 1 by definition. So, take 0 along with all sums of the form 1, 1+1, 1+1+1 and so on. We can call this set N. We can take Z to be the set of all elements of N and all additive inverses of elements of N. Finally take Q to be the set containing all elements of Z and all multiplicative inverses of (nonzero) elements of Z. Now we have our R and Q. Also, each step of the above follows from our field axioms. Defining irrationals is straightforward from this.
So, the definition bit here is not a problem. The bigger issue is that this definition doesn’t tell us that a complete ordered field exists. We can define things that don’t exist, like purple flying pigs and so on.
What the dedekind cut construction shows is that using only the axioms of zfc we can construct at least one complete ordered field.
For reference:
https://en.m.wikipedia.org/wiki/Real_numbers
https://en.m.wikipedia.org/wiki/Ordered_field