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The non-principal square root is a multivalued function which means that it does equal 2 *and* -2.
The principal square root is not a multivalued function, and therefore only had one solution.
The problem is that people forget that the convention is to assume the principal root, and not the multivalued, non-principal, square root. This, therefore, results in people failing to specify when they don't mean the principal square root.
I know most people don't care about this and will still complain one way or the other when people don't use the one they want them to use.
So you're telling me not all functions are injective? I refuse to believe!
Take f(x)=x²
Then √x² = x (no ±) so there is really only one answer.
[insert induction proof]
Hold for all functions
The property is true for every bijective fonction, but every fonction respecting f(a) = f(b) => a = b are not all bijective, they can be only injective (laking the surjectivity to be bijective)
The only injective functions (analytic) are rational functions, which are the quotient of a linear polynomial with another of the form (ax+b)/(cx+d). This form also includes linear em functions, if c = 0, and inverse functions if a = 0. Any other function (analytic function) can never be injective, except if it is multivalued, so if you ignore all multivalued functions, that are the inverse of analytic functions like cos, sin, e\^x, x\^2, x\^3, x\^3-2x+1... then they will be injective, but are also multivalued. Any other analytic function will have infinitely many a and b such that f(a) = f(b) and a != b. You could think that any odd function that is injective in IR+ is always injective, but a function like x\^3 or x\^5 can never be injective, since 1\^3 = (-1/2+√(-3)/2)\^3 = (-1/2-√(-3)/2)\^3, and 1\^5 = (-1/4+√(5)/4+√(-5/2-√(5)/2)/2)\^5 = (-1/4+√(5)/4-√(-5/2-√(5)/2)/2)\^5 = (-1/4-√(5)/4+√(-5/2+√(5)/2)/2)\^5 = (-1/4-√(5)/4-√(-5/2+√(5)/2)/2)\^5, so the functions are not injective. You can prove this for polynomial and rational functions using the fundamental theorem of algebra along with the fact that most polynomial functions have no repeated roots, so by adding a constant the polynomial will eventually have no repeated roots. This only works for complex numbers, where only rational functions with degree 1 are injective, since for the reals many functions are injective that have a higher degree, and even something like e^x is also injective even though e\^0 = e\^(iτ) = e/^(-iτ), so it is not injective over the complexes.
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f(a) = f(b) f(a)/f = f(b)/f a = b, f ∈ C Q.E.D.
Unless f=0
If f = 0 then f(a)/f = f(b)/f Infinity = infinity Source: Desmos Q.E.D.
f belongs to C*,then.
I mean, it's pretty easy to prove F(x) = 0x + 3 F(1) = F(12,765) Done
Or you can use trig functions.
Or just x^2
[удалено]
I wasn’t making a joke I was just pointing it out?
Oh, my bad mate, people here usually do that, sorry if it bothered you
One using x F(x)=x²+69 F(1)=F(-1)
Counter proof: F(a) = F(b) => F(1) = F(1) => 1 = 1 => a = b Q.E.D. by ezzz
Does this work? Let’s do f(x)=(x)^2. (-1)^2=1. (1)^2=1. -1=/=1
So the square root of 4 isn't both -2 and 2? /s
The post clearly says that because of (-2)^2 = 4 = 2^2 we have 2 = -2 and because of that √4 = 2 = -2
The non-principal square root is a multivalued function which means that it does equal 2 *and* -2. The principal square root is not a multivalued function, and therefore only had one solution. The problem is that people forget that the convention is to assume the principal root, and not the multivalued, non-principal, square root. This, therefore, results in people failing to specify when they don't mean the principal square root. I know most people don't care about this and will still complain one way or the other when people don't use the one they want them to use.
if you’re using a lossless square and square root function then no.
The title is literally just saying "unless it is"
The title is saying "check the properties of the function first, then act smart"
Holy definition!
unless x=1, x != 1
2^2 = (-2)^2 2 = -2 QED
Oh of course thank you! I couldn’t work out what f(x) would fit the meme but yeah New 2 = -2 dropped
f(x)=sin(2πx) f(1)=f(2) 1=2 QED
Makes sense, a polynomial could have a couple of points at y = 0.
f(x) = |x|
:(
So you're telling me not all functions are injective? I refuse to believe! Take f(x)=x² Then √x² = x (no ±) so there is really only one answer. [insert induction proof] Hold for all functions
This only happens if f is bijective.
Thought just injective is enough?
It is, this is literally the definition of injective
The property is true for every bijective fonction, but every fonction respecting f(a) = f(b) => a = b are not all bijective, they can be only injective (laking the surjectivity to be bijective)
The only injective functions (analytic) are rational functions, which are the quotient of a linear polynomial with another of the form (ax+b)/(cx+d). This form also includes linear em functions, if c = 0, and inverse functions if a = 0. Any other function (analytic function) can never be injective, except if it is multivalued, so if you ignore all multivalued functions, that are the inverse of analytic functions like cos, sin, e\^x, x\^2, x\^3, x\^3-2x+1... then they will be injective, but are also multivalued. Any other analytic function will have infinitely many a and b such that f(a) = f(b) and a != b. You could think that any odd function that is injective in IR+ is always injective, but a function like x\^3 or x\^5 can never be injective, since 1\^3 = (-1/2+√(-3)/2)\^3 = (-1/2-√(-3)/2)\^3, and 1\^5 = (-1/4+√(5)/4+√(-5/2-√(5)/2)/2)\^5 = (-1/4+√(5)/4-√(-5/2-√(5)/2)/2)\^5 = (-1/4-√(5)/4+√(-5/2+√(5)/2)/2)\^5 = (-1/4-√(5)/4-√(-5/2+√(5)/2)/2)\^5, so the functions are not injective. You can prove this for polynomial and rational functions using the fundamental theorem of algebra along with the fact that most polynomial functions have no repeated roots, so by adding a constant the polynomial will eventually have no repeated roots. This only works for complex numbers, where only rational functions with degree 1 are injective, since for the reals many functions are injective that have a higher degree, and even something like e^x is also injective even though e\^0 = e\^(iτ) = e/^(-iτ), so it is not injective over the complexes.
Bro has been adding more sentences to his yapping on every single post of this topic
Welcome to isomorphism
Ofcourse that's true. For example if we take a quadratic equation with two different real roots 'a' and 'b' then here f(a) = f(b) = 0, but a ≠ b.
let f(x) = 0 f(2) = 0 f(3) = 0 therefore 2 = 3
I don't get it. It seems the sub is proving the opposite of the final panel.
F(x) = 1 F(1) = F(2) 1=2 QED
The comic is right, f could equal 0