/sci/ - Science & Math

[Anonymous No.10939066]

(-1)^0 = 1

I'm an idiot, why is this the case?

[Anonymous No.10939095]

>>10939066
>I'm an idiot, why is this the case?
(-1)^0 =
(-1)^(1-1) =
(-1)^(1)/(-1)^(1) =
1/1 =
1

[Anonymous No.10939126]

>>10939066
a^b === e^(b*log(a))

-1^0 = e^(0*log(-1))
=e^0
=1

[Anonymous No.10939130]

>>10939095
>>10939126
>make up a bunch of arbitarry rules
>bro it's logical, lol
Mathematicians are just delusional physicists. At least the latter admit for their research to be poorly concocted bullshit.

[Anonymous No.10939139]

>>10939066
(for any real number)^0=1

[Anonymous No.10939140]

>>10939066
Because
[eqn] -1 = \exp [i\pi] [/eqn]

[Anonymous No.10939145]

>>10939126
>-1^0 = e^(0*log(-1))
>log(-1)
u fkn wt

[Anonymous No.10939165]

>>10939139
Which tells the anon nothing of why or what or how that is.

[Anonymous No.10939190]

(-1)^0=1
(-1)^1=-1

[WHAT THE FUCK IS GOING ON No.10939196]

(-1)^0=1
(-1)^1=-1
(-1)^2=1
(-1)^3=-1
(-1)^4=1
(-1)^5=-1

[Anonymous No.10939198]

>>10939130
I mean how are the arbitrary? They are the rules of exponentiation without any exceptions.

[Anonymous No.10939200]

>>10939130
Shut up retard

[Anonymous No.10939208]

>>10939198
>rules of exponentiation
So rules that are part of a bigger chunk of arbitary rules.

[Anonymous No.10939221]

>>10939190
>>10939196
(-1)^x = cos(pi*x) + i*sin(pi*x)

[Anonymous No.10939242]

I still don't understand

(-1)^0= 1 * (-1) = -1 because a negative * a positive should = a negative

What am I missing?

[Anonymous No.10939263]

>>10939242
>(-1)^0= 1 * (-1)
How on earth did you come up with that conclusion?

[Anonymous No.10939290]

>>10939242
It's not what you're missing, but what you're adding.

n^0 implies no multiplications by n. The final equation should look like:

[math]1[/math]

[Anonymous No.10939306]

>>10939145
>He doesn't know compkex logs and branches
Lol

[Anonymous No.10939312]

>>10939066
sneed

[Anonymous No.10939313]

>>10939306
;-;

[Anonymous No.10939568]

>>10939165
To be honest, this >>10939139 makes more sense than this >>10939095

It's much easier to say "because it is" than to accept some ridiculous math that just happens to work out. Can you give me an actual worded explanation as to why (-1)^0=1

[Anonymous No.10939583]

>>10939066
(-1)^4=1
(-1)^3=-1
(-1)^2=1
(-1)^1=-1
(-1)^0=1
(-1)^-1=-1
(-1)^-2=1
(-1)^-3=-1
(-1)^-4=1

[Anonymous No.10939590]

[math]
x^n.x = x^(n+1) so x^n = x^(n+1)/n

x^2 = x^3 / x = x^2
x^1 = x^2 / x = x
x^0 = x^1 / x = 1
x^-1 = x^0 / x = 1/x
...
[/math]

[Anonymous No.10939594]

[math]x^n.x = x^(n+1) so x^n = x^(n+1)/n[/math]

[math]x^2 = x^3 / x = x^2[/math]
[math]x^1 = x^2 / x = x[/math]
[math]x^0 = x^1 / x = 1[/math]
[math]x^-1 = x^0 / x = 1/x[/math]

[Anonymous No.10939597]

[math]x^n.x = x^{n+1}[/math]
so
[math]x^n = x^{n+1}/n[/math]

[math]x^2 = x^3 / x = x^2[/math]
[math]x^1 = x^2 / x = x[/math]
[math]x^0 = x^1 / x = 1[/math]
[math]x^(-1) = x^0 / x = 1/x[/math]

[Anonymous No.10939605]

[math]x^n.x = x^{n+1}[/math]
so
[math]x^n = x^{n+1}/n[/math]

[math]x^2 = x^3 / x = x^2[/math]
[math]x^1 = x^2 / x = x[/math]
[math]x^0 = x^1 / x = 1[/math]
[math]x^{-1} = x^0 / x = 1/x[/math]

[Anonymous No.10940757]

>>10939130
>arbitarry rules
>arbitarry
Lrn2arbitarry fgt pls

[Anonymous No.10940759]

OP let's start with simple things:
[math]a^{1} = a x 1 = a, thus -1^{1} = -1[\math] right?

Using basic multiplication properties we have the following:

[math]-1^{1} / -1^{1} = -1^{1-1} = -1^0 [\math]

and
we know that a number multiplied by its reciprocal equals the neutral element of multiplication, thus:

[math] -1^{1} / -1^{1} = 1.[\math]

Using both equations we have that:
[math] 1= -1^{1}/-1^{1}= -1^{0} [\math]

[Anonymous No.10940763]

>>10939208
>arbitary rules
...is Tuesday the official "Arbitrary Rules Faggot Day"??

[Anonymous No.10940764]

>>10940759
Fuck

[Anonymous No.10940771]

>>10939139
0

[Anonymous No.10940773]

>>10939208
>arbitary rules
>arbitary
Lrn2arbitary fgt pls

[Anonymous No.10940835]

>>10939066

This is another one of those "the math checks out, so it's true!"

Even though this literally makes no sense and has no real world analogue and wouldn't work if there even could be a real world analogue.

[Anonymous No.10940843]

>>10939066
It's useful.
Special cases are for fags.
YOLO

[Anonymous No.10941061]

>>10940835
>>10939568
>>10939066
don't call something "ridiculous math that just happens to work out" if you don't know what you're talking about. also don't say "it makes no sense and has no real world analogue" if you don't know what you're talking about.

algebraic reason:
We have [math]x^a[/math] defined for [math]a[/math] being a positive integer, it's just repeated multiplication. It's easy to see that for positive integral exponents we have [math]x^{a+b} = x^ax^b[/math]. We would now like to define exponentation also for zero and negative exponents. We want this extension to be somehow meaningful: if possible we want the same algebraic properties. So if we want the identity [math]x^mx^n = x^{m+n}[/math] to remain true for arbitrary exponents, we necessarily have [math]x^0 = x^{0+0} = x^0x^0[/math] which implies that [math]x^0 = 1[/math]. We also must have [math]x^{-a}x^{a}=x^{a-a}=x^0=1[/math] and this implies [math]x^{-a} = \tfrac{1}{x^a}[/math]. This tells us how exponentation for negative and zero exponents MUST be defined in order to stay consistent, there's no other choice. if we define it this way, it's easy to check that these rules indeed stay valid.

[Anonymous No.10941068]

>>10940835
>>10939568
>>10939066

geometric reason:
We can take numbers to represent transformation: take your favorite object (a car or a banana or whatever) and each number represents a scaling of the object. [math]3[/math] means make it three-times larger, [math]\tfrac{1}{3}[/math] means make it three times smaller. [math]-3[/math] means make it three times larger and also reflect it along some fixed axis or plane. multiplication now represents composition of these scalings: if you make the banana two times larger and then three times larger , it will get six time larger than the original size etc. [math]1[/math] is a neutral element, because it literally says "do nothing". [math]1=3*\tfrac{1}{3}[/math] means that if you make it three times larger and then three times smaller, nothing happens. [math]-1[/math] represents just the reflection, without changing size. clearly [math](-1)*(-1) = 1[/math] says that two reflections cancel each other out. etc. etc. you see that everything checks out. we define [math]x^k[/math] to be: for [math]k>0[/math] it means do the scaling [math]x[/math] succesively [math]k[/math]-times, for [math]k<0[/math] it means do the [math]\tfrac{1}{x}[/math] scaling succesively [math]-k[/math]-times. you can easily check that the usual arithmetics for exponentation like [math]x^ax^b = x^{a+b}[/math] now hold for these purely geometric reasons. as for the zero, [math]x^0[/math] should literally mean "do [math]x[/math] zero times", which means "do nothing". and doing nothing is [math]1[/math].

[Anonymous No.10941071]

>>10939139
>what is 0^0

[Anonymous No.10941089]

>>10941071
>>10940771
Brainlets

[Anonymous No.10941098]

>>10941089
>Zero to the power of zero, denoted by 0^0, is a mathematical expression with no agreed-upon value.

[Anonymous No.10941119]

>>10941071
1

[Anonymous No.10941123]

>>10941098
It’s actually almost always agreed to be 1. If you don’t accept that convention, you have to make special cases for the binomial theorem, the connection between powers and number of functions between some domain and codomain, etc.

TLDR 0^0 = 1

[Anonymous No.10941124]

[math]-1 = e^{i\pi}[/math]
therefore
[math]-1^x = e^{i\pi x}[/math]

plug in a zero for x, what do you get? 1

[Anonymous No.10941127]

>>10941098
it’s literally the empty product (i.e. 1) you stupid idiot

[Anonymous No.10941132]

>>10941127

I see. So you guys are just making shit up because it LITERALLY cannot equal 1. 0 groups of 0?

Bruh..........

Get off that dogma shit.

[Anonymous No.10941135]

>>10941127
>https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero
Retard.

[Anonymous No.10941147]

>>10941135
Dumb retard did you read the fucking article you linked to? There are very good reasons to define it as 1 and NO food reasons not to.

[Anonymous No.10941153]

>>10941147
Nice Freudian slip there, you fat fuck

[Anonymous No.10941156]

>>10941132
Zero groups of zero is additively zero.

Zero groups of zero is multiplicatively one.

[Anonymous No.10941157]

>>10941153
>latches onto a typo and ignores the rest
>if you ever mention food you must be fat
Thanks for confirming that you’re mentally handicapped

[Anonymous No.10941162]

>>10941132
what the fuck are you even saying

[Anonymous No.10941163]

>>10941157
I'm qualified for Mensa actually. Do you really think Wikipedia would state that there was no agreed upon definition for 0^0 if 0^0 NECESSARILY meant 1?

[Anonymous No.10941167]

>>10941132
It can equal whatever you want, we are setting the definition to be useful to us.

For example, in some instances 0! = 1 and in some instances 0! = 0
It's basically just whatever we find useful at the time

[Anonymous No.10941181]

>>10939208
They aren't arbitrary but they are made up

[Anonymous No.10941206]

>>10941163
>I’m qualified for Mensa actually.
>adding even more confirmation that you are indeed mentally handicapped.

[Anonymous No.10941211]

>>10941206
Retarded cunt. Show me your IQ then, faggot.

[Anonymous No.10941222]

>>10939066
>>10939095
>>10939126
>>10939130
>>10939139
>>10939140
>>10939145
>>10939165
>>10939190
>>10939196
>>10939198
>>10939200
>>10939208
>>10939221
>>10939242
>>10939263
>>10939290
>>10939306
@10939312
>>10939313
>>10939568
>>10939583
>>10939605
>>10940757
>>10940759
>>10940763
>>10940764
>>10940771
>>10940773
>>10940835
>>10940843
>>10941061
>>10941068
>>10941071
>>10941089
>>10941098
>>10941119
>>10941123
>>10941124
>>10941127
>>10941132
>>10941135
>>10941147
>>10941153
>>10941153
>>10941156
>>10941157
>>10941162
>>10941163
In any commutative monoid, the operation on zero elements (i.e. the empty operation) is defined to be the identity element of the monoid.
Examples:
empty sum = zero
empty product = one
empty max = -infinity
empty min = +infinity
empty conjunction = true
empty disjunction = false
empty xor = false
empty lcm = one
and so on. The reason for this is simple: Let a and b be lists of elements of the monoid and || denote concatenation of lists. In the case of the sum, for example, you want

sum (a || b) = sum a + sum b

Thus if a is empty you have a || b = b so you want

sum b = sum a + sum b

so sum a must be the identity with respect to +. Same goes for product and multiplication.

[Anonymous No.10941224]

>>10941167
I have never ever seen a situation where 0! = 0 makes sense. 0! = 1 on the other hand always makes sense.

[Anonymous No.10941240]

>>10941156

Multiplication is literally just addition quickly.

[Anonymous No.10941244]

>>10941240
Yeah, if you are in first grade maybe

[Anonymous No.10941247]

>>10941167

So you agree that it's all abitrary then. That's totally fine, just admit that some things are just made up for convenience because the math balances out.

Because there is no fucking reason why anyone would think that 0^0=1 or why (-1)^0 = 1. And giving mathematical proofs that show that 0^0=1 is not "proving" it.

[Anonymous No.10941251]

>>10941244

Please show me an instance where it isn't.

[Anonymous No.10941264]

>>10941247
>Because there is no fucking reason why anyone would think that 0^0=1 or why (-1)^0 = 1
it’s almost like you didn’t read the thread retard >>10941222

[Anonymous No.10941268]

>>10941251
Multiplication by -1.
Multiplication by 1/2.
Multiplication by sqrt 2.
Multiplication by pi.
Multiplication by i.
Need more examples?

[Anonymous No.10941271]

>>10941251
97*103
= (100-3)(100+3)
= 100^2 - 3^2
= 10,000 - 9
= 9,991.

Not gonna find any elementary school fucks doing this.

[Anonymous No.10941273]

>>10941240
The structure of a group is matter of definition, not set composition.

[Anonymous No.10941280]

>>10941251
[math]\left(\frac{\sqrt{\pi}}{2}+ie\right)\cdot\left(\left(\frac{2}{3\pi}\right)^{4/3}-7i\right)[\math]

What is the addition here?

[Anonymous No.10941282]

>>10941251
>>10941280
[math]\left(\frac{\sqrt{\pi}}{2}+ie\right)\cdot\left(\left(\frac{2}{3\pi}\right)^{4/3}-7i\right)[/math]

Sorry, a typo

[Anonymous No.10941285]

>>10939568
Which step did you find "ridiculous", just out of curiosity?

[Anonymous No.10941290]

>>10941167
>in some instances 0! = 0
No.

[Anonymous No.10941294]

>>10941285
I didn't read that post as adversarial in tone.

[Anonymous No.10941323]

>>10941247
>giving mathematical proofs that show that 0^0=1 is not "proving" it

[Anonymous No.10941342]

>>10939066
^ is a subdomain under *.

[Anonymous No.10941375]

>>10941271

You solved it by not using addition, but it's literally just 97 groups of 103.

>>10941268
4 x pi
Just four groups of pi. Just press the Pi button four times on your calculator, bro.

[Anonymous No.10941380]

>>10941323

It just shows that the "math works." Not that it actually makes sense in any sort of physical capacity. Perhaps we don't have a good enough mathematical system to explain how these answers can possibly be.

[Anonymous No.10941383]

>>10941375
What about pi x pi? Good luck pressing pi pi times on your calculator

[Anonymous No.10941439]

>>10941383
Is the amount of groups a discrete or continuous function?

[Anonymous No.10941472]

>>10941380
Let me guess: You're one of those faggots who think negative numbers, rationals, irrationals, etc. do not "make sense in any sort of physical capacity" despite them underlying our best theories of how the world works.
tl;dr kys fag

[Anonymous No.10941487]

>>10939066
-1^1 = -1
-1^-1 = 1/-1 = -1
-1^-1*-1^1 = -1^(-1+1) = -1^0 = -1*-1 = 1
>>10940759
cumbersome but right.
Everyone else failed hard.

[Anonymous No.10941540]

Mate, check it:

[math] 1 = \frac{-1^1}{-1^1} = -1^1 \cdot \frac{1}{-1^1} = -1^1 \cdot -1^{-1} = -1^{1+(-1)} = -1^0 [/math]

ezpz