Not trolling, though I'm sure this is an easy question for the mathematically inclined

the Geneva convention?

explain to me how to imagine 2^0.1

I can visualize 2^2=2*2 but the first thing is so vague to me i cant grasp it

No. It's the only logical way to interpret it. Sorry if your brainlet teachers told you otherwise.

>Wrong. -2^2 = (-2)*(-2) by convention.
incorrect

-2^2 = -1*2^2 = -1*2*2

Honest question for you guys.
Do you actually use pemdas when you're doing your hw and tests?
I've always established -2^2 = (-2)^2. I've literally never seen it any other way outside of these threads. I've never seen professors take -2^2 as -(2^2) either.
Similarly, what is 2^(1/2)?

It has higher precedence then multiplication
[math]
2x^2 = 2 \cdot x \cdot x
[/math]
and not
[math]
2x^2 = 2 \cdot x \cdot 2 \cdot x
[/math]
So:
[math]
-2^2 = (-1)\cdot 2^2 = (-1) \cdot 4 = -4
[/math]

As a fraction:
[math]
1 = \frac{2^a}{2^a} = 2^{a-a} = 2^0
[/math]
This also makes immediately clear why [math] 0^0 [/math] is not defined.

No, I've never ever seen it this way.
[math] 2^{1/2} = \sqrt{2} [/math]
or what do you mean?

not sure what you mean, not even sure what pemdas really stand for, but yes you always use, thats how mathematics work
its possible some shitty textbooks have mistakes in them but -2^2 is always (-1) * 2*2

>Similarly, what is 2^(1/2)?
[math]\sqrt[n]{{X}^{m}}=X^\tfrac{m}{n}[/math]

>2^.1 = 2^(1/10) = 2^1*2^(-10) = 1/2^9 = 1/512
hahahahahahahahahahahhahahahahaah
do you even math?

First realize that it has to be that way or else it wouldn't work

What is the sign of the value of sqrt(2)?

Commonly, -x is treated as its own object. Then -x^2 = x^2 implies (-x)^2. It's an abuse of notation, but it's still valid.

>replying to a troll thread
I remember the days when trolls didn't have to annouce "i'm not trolling", before trolling. Today's trolls are weak dumb faggots.

>Commonly, -x is treated as its own object
maybe by laymen like yourself
-x is -1 * (x)

0-2^2 = ?

en.wikipedia.org/wiki/Unary_operation

Hopefully this settles it.

infix is truly the cruelest, most malignant, metastasized cancer afflicting mathematics, after the so-called real numbers

2*2 = 4
(-2)^2 = 4

So (4)^(1/2) = +/- 2

However, if you use the square root symbol:
sqrt(4) = 2
because that symbol specifically talks about the PRINCIPLE square root (meaning the positive root)

...

if i remember right, sqrt(2) has two solutions, -2 and 2. but the notation with the radical sign or a fractional exponent refers only to the principal (positive) root.

What is the point of the term i such as 4i^2=-4?
Serious question.

"i" signifies an imaginary number.

Here is why it is is used:

sqrt4=2 or -2

because 2*2=4 and -2*-2=4

Notice that you can't achieve a negative sqrt.

sqrt(-4) is not 2 or -2, because again squaring those yields a positive 4.

so 4i is how you right the sqrt of -4.

so if 4i =sqrt-4 Then squaring that number 4i^2 would remove the sqrt from the -4, thus you are left with -4.

TL;DR:
the "i" describes the square root of a negative number.

Okay thank you, I actually learned something on Veeky Forums.

Your welcome.
this can also happen

sqrt-8 =2sqrt2i

Brainlet

Don't listen to them.
Sqrt (4) is 2 and only 2.
x^2 = 4 has solutions x = +-2.

If you write sqrt (x^2) = +- x you'll be docked for it at uni.

the proper sqrt of 4 is 2
the actual solution for x^2 = 4 is both +-2 like you said
uni proffs are just dumb and think the proper sqrt is better

Nearly, but not quite. i=sqrt(-1), so in your example with -4, sqrt(-4)=sqrt(-1*4)=sqrt(-1)*sqrt(4)=i*2=2i.
Not 4i

you are a fucken retard

stop responding to bait

Wrong. No such convention exists. -a == -1*a,
order of operations to conclude -a^2 = -1 * a^2

-2^2 is -2 * -2 by definition, retard. Get back in to middle school

No. It is not. See

What retard tier degree did you take where the proffs where that useless at maths?

Clearly (because precedence):
[math]
0-2^2 = -4
[/math]
and
[math]-a^2 = -a \cdot a[/math]
And normally:
[math]-2^2 = -4[/math]
But you can interpret [math] -2 \in \mathbb{Z} [/math] which would indeed leed to [math] -2^2 = 4 [/math] (to be honest I've never seen this but it is legit)

How the fuck else in your Wikipedia self taught maths land do you represent a negative number other than putting minus sign infront of

>This also makes immediately clear why [math]0^0[/math] is not defined
[math]0^0 = 1[/math] by convention

As (-2):
Add a negative number: 5+(-2)
Substract a positive number: 5-2

wolframalpha.com/input/?i=0^0

Most people who write maths on paper rather than in a computer will save themselves the effort of writing all those unneeded brackets and write (-2) as -2. There is no difference between adding a negative number and subtracting a possitive number so those brackets are not needed.
Brackets are generally only used when they are needed to avoid confusion in a expression. Most people who see (-2^2) would read it as ((-2)^2) by your reading.

However, this doesn't address why you think scrt(4)=/=+-2.

I know this is pedantic, but technically they are not the same. One is over [math] \mathbb{Z} [/math] the other one over [math] \mathbb{N} [/math],
>Most people who see (-2^2) would read it as ((-2)^2) by your reading.
Yes, and as I said, I've never seen it that way. All I say is that one should be cautious about ambiguity and that the "-2^2 = 4"-group has an argument.

>However, this doesn't address why you think scrt(4)=/=+-2.
? I didn't post anything concerning the sqrt-debate. But it's
[math] \sqrt{4} = \pm 2 [/math], as an operation and
[math] \sqrt{4} = 2 [/math], as an function.
So, ambiguity again...

Sqrt (x^2) = |x|
Sqrt is a function, not an operation.

The only way you get two solutions is if you equate it.
y = sqrt (x^2) = |x| so y = +-x
Without an operator, you can't get two values from a function. The fact that you don't know this tells me you're the one going to a shit uni.

Furthermore, -x should strictly be interpreted as the unary operator with x as its argument. Unary operations take precedence over anything else.

Note that -2 is not (-1*2), it is equivalent to the binary operator * with arguments -1 and 2. Similarly, 0-2 is a binary operator - with 0 and 2 as arguments and is equivalent to -2.

Since they are equivalent, you must evaluate the equivalency before any other operations are performed.
-2^2 = (0-2)^2 = (-1*2)^2

In all cases, -2^2 = 4
This is a formal definition, but people often abuse the notation to make troll threads like these.

Just like a differential operator d/dx, we often cross multiply terms as if the d/dx was a fraction, but this is completely fictitious and only works because it is equivalent to successive applications of the operator.

Now please go ask your proff why he hadn't told you about this yet.

Just to be absolutely solid that you're wrong. Consider the following:
-a^2
-a = xa for some integer x

Then -a^2 = (xa)^2.

In your argument, you claim -a^2 = (xa)^2 = x (a)^2

This is absurd.

As another example:
-a^2 evaluate
-a = (bd + cd)
It does not follow that
(bd+cd)^2 = d (b+c)^2
But that's what anyone whow makes your argument is doing.

>all these pompous morons getting baited by a troll.
come on guys. this is as bad as people saying .999...=/=1
Why even fall for this

There's enough people in my major, and in everyday life that genuinely believe these things. I doubt all of the posts saying otherwise were trolling.

>46 replies and 3 images omitted
brainlet thread detected

-a^2 =/= (-1)^2 * a^2
brackets

[math]2^{0,1} = 2^{\frac{1}{10}} = \sqrt[10]{2^1}[/math]
Well basically you can't multiply it 0.1 times so instead we just extract a root

What about [math]2^{\pi}[/math]?

sqrt(2) is approximated by 1.414 because it is an irrational

It sure does.
-a = -1*a
So -a^2 = (-1*a)^2
You can distribute the exponent here:
= (-1)^2 (a)^2 = 1*a^2 = a^2.

[eqn]2^\pi = \lim_{n \to \infty}2^{x_n}[/eqn]
where [math](x_n) [/math] is any sequence of rational numbers that converges against [math]\pi [/math].

>unironically using the axiom of infinity
Cmon bruh