Mathematical Curiosities. (3 Viewers)

[anomalousdecay]

No, not when i means what it normally does.

Ok thanks. I'm guessing the "not nice" complex numbers are above my pay grade lol.
Something I may come across in uni.

 

[seanieg89]

Ok thanks. I'm guessing the "not nice" complex numbers are above my pay grade lol.
Something I may come across in uni.

Well they wouldn't be called "complex numbers" anymore. Yep, depending on what you do, you might.

 

[anomalousdecay]

Well they wouldn't be called "complex numbers" anymore. Yep, depending on what you do, you might.

Engineering. However, I may do a double with a B Sci in either Maths or Chemistry.

EDIT:

Here is another odd derivation for i, which is found in Terry Lee's textbook iirc.







I've always wondered why the second line is false.



The actual proof:









Is this a property of square roots that we just simply "neglect" for real numbers?

 

[asianese]

Double sci engin is a tough degree. Long.

 

[anomalousdecay]

Double sci engin is a tough degree. Long.

Isn't it 5 years at UNSW? Any way, off topic here, I made a thread not too long ago about that here:

http://community.boredofstudies.org/showthread.php?t=314852

 

[RealiseNothing]

I would also like to know. All I found for it was https://en.wikipedia.org/wiki/Quaternion (see the table on the right). "i^2 = j^2 = k^2 = ijk = −1,"

You should have a read of the story behind how Hamilton came up with that idea, it's crazy.

 

[seanieg89]

Engineering. However, I may do a double with a B Sci in either Maths or Chemistry.

EDIT:

Here is another odd derivation for i, which is found in Terry Lee's textbook iirc.







I've always wondered why the second line is false.



The actual proof:









Is this a property of square roots that we just simply "neglect" for real numbers?

Firstly, you shouldn't expect too much rigour in high school. Constructing the complex numbers is a little more subtle than just saying:

"Let's assume that there exists a square root of -1 and call it i."

In this case, the flaw is that sqrt(ab) is not necessarily equal to sqrt(a)sqrt(b) for any single-valued definition of the square root function. (such as the principal square root). That this rule is valid for positive real numbers and the sqrt function defined on them is no indication that the rule should be valid when the function is extended a larger set.

 

[anomalousdecay]

Firstly, you shouldn't expect too much rigour in high school. Constructing the complex numbers is a little more subtle than just saying:

"Let's assume that there exists a square root of -1 and call it i."

In this case, the flaw is that sqrt(ab) is not necessarily equal to sqrt(a)sqrt(b) for any single-valued definition of the square root function. (such as the principal square root). That this rule is valid for positive real numbers and the sqrt function defined on them is no indication that the rule should be valid when the function is extended a larger set.

So this is a limit to the shortcuts in complex numbers I guess.

Wow. Complex Numbers are so fascinating. The stuff we learnt in 4U about complex numbers is barely an introduction then.

 

[seanieg89]

So this is a limit to the shortcuts in complex numbers I guess.

Wow. Complex Numbers are so fascinating. The stuff we learnt in 4U about complex numbers is barely an introduction then.

Yeah, if you ever learn any complex analysis or analytic number theory your mind will be blown. Staggeringly beautiful subject.

 

[anomalousdecay]

Yeah, if you ever learn any complex analysis or analytic number theory your mind will be blown. Staggeringly beautiful subject.

The complexity and abstract arguments in maths make it so beautiful. In essence, it is a language that few understand well.

 

[asianese]

lol that sounds so wanky. (no offence)

 

[anomalousdecay]

lol that sounds so wanky. (no offence)

What has English advanced done to me?

 

[asianese]

Everything undesirable.

 

[anomalousdecay]

Everything undesirable.

What has maths done to me?

Everything desirable.

 

[HeroicPandas]

Why is a^m+n = a^m x aⁿ? (and the rest of the index laws, beside a^0 = 1)

i've always remember index laws but never proved it

 

[anomalousdecay]

Why is a^m+n = a^m x aⁿ? (and the rest of the index laws, beside a^0 = 1)

i've always remember index laws but never proved it

So we have multiplied a by itself m times and then n times, in separate brackets. Together we get an integer, which we shall call q.




Again, we have a multiplied by itself m+n times, so we get q.


Its really hard to understand the proof, but if you look at it closely, you should get the result.
I need someone's help explaining this proof a little more.

 

[RealiseNothing]

Why is a^m+n = a^m x aⁿ? (and the rest of the index laws, beside a^0 = 1)

i've always remember index laws but never proved it

You have 'm' lots of 'a' times by 'n' lots of 'a'. So altogether you have 'm+n' lots of 'a'.

 

[seanieg89]

You have 'm' lots of 'a' times by 'n' lots of 'a'. So altogether you have 'm+n' lots of 'a'.

Yep, of course the formal proof would proceed by something like induction, using that multiplication of reals is associative. Such a proof is terribly boring though and doesn't tell you anything that is at all surprising.

Note though, that even this formal proof would only show that a^{m+n}=a^ma^n for all non-negative integers m,n. You can extend this to integers and rationals as well quite easily.

For arbitrary real/complex exponents however, this would not be valid, and we would need to refer to how we define such powers.

 

[asianese]

We can refer to Dave easdowns inquisitive interpretation of exponentials heh

 

[HeroicPandas]

So we have multiplied a by itself m times and then n times, in separate brackets. Together we get an integer, which we shall call q.




Again, we have a multiplied by itself m+n times, so we get q.


Its really hard to understand the proof, but if you look at it closely, you should get the result.
I need someone's help explaining this proof a little more.

You have 'm' lots of 'a' times by 'n' lots of 'a'. So altogether you have 'm+n' lots of 'a'.

thanks i understand