Let's say I need to prove that the expression of a function f(x) can be re-written in some other way than its original expression.
To do that, can I just assume and state that both expressions of the function are equal, and then see how it goes from there? So that if I will arrive at an equality with same expressions on each hand, the proof will be completed. (If the opposite is true, though, it would prove that the function can't be re-written as that expression)
Is there a faster and more acceptable way to do proofs like that?
>>8060942
Proof that the thing that is changed is the same can sometimes be faster. Can u give an example?
>>8060949
This is a very basic example that I came up with:
Prove that the square root of x^3 can be rewritten as x^(3/2).
What I do is write sqrt(x^3) = x^(3/2), and then work from there.
>>8060958
Why would you need to proof this? These are two times literally the same function. Wtf anon?
>>8060958
Well if you have sqrt(x^5 + x^3) ^3
You can just say f(x) =x^5 + x^3
And so on...
>>8060962
They are. And the goal of the proof is to prove that those two functions are the same. Is setting both sides as equal and doing algebra until one arrives at the same expressions on both sides the fastest way? This is what I've always done, but I'm just wondering.
>>8060994
Ex:
sqrt(x^3) = x^(3/2)
Then doing algebra, until one arrives to the points where both sides are identical.
>>8060998
They already are the same. Looking the same has no mathematical value.
It's null property
>>8060942
If A = B then A - B = B - A = 0
>>8060942
No, you cannot do that I and would take points off of homework when students did that when I was grading for an intro to proof writing course.
You cannot assume that what you're trying to prove is true.
The correct way to do this is to start with one expression, and manipulation it until you find the other side.
An extremely simple example:
Prove that (x+2)^2 = x^2 + 4(x+1).
Proof:
(x+2)^2=(x+2)(x+2)
= x^2+2x+2x+4
= x^2 +4x+4
= x^2 +4(x+1).
You should never do it like this:
(x+2)^2=x^2+4(x+1)
(x+2)(x+2)=x^2+4x+4
x^2+4x+4=x^2+4x+4
>>8061046
But why can't you assume that it's true and then attempt to derive a contradiction? If no contradiction is derived, doesn't it mean that the statement is correct?
>>8061018
Going by that logic, all math has no mathematical value. It's just a different way to state the same definitions, axioms and rules of inference.
>>8061046
Here's an example of why you're wrong.
Let's take
f(x) = x^2+3
and
g(x) = x^3+7
If we set x^2+3 = x^3+7, it is clear that we will never arrive at the same expression on both sides. Therefore, the expressions aren't expressing the same function at all:
x^2 = x^3+10
x = sqrt(x^3+10)
x = (x^3+10)^(1/2)
It is clear that this equality doesn't hold, so the expressions don't denote the same function.
On the other hand, let's take
f(x) = sqrt(x^3)
g(x) = x^(3/2)
Here, both functions are the same.
We can prove this by this reasoning:
If both, f(x) and g(x) are the same functions, we would arrive at the same expression on both sides when we set f(x) = g(x).
And we indeed will do so:
sqrt(x^3) = x^(3/2)
sqrt(x^3) = x^(3*(1/2))
sqrt(x^3) = sqrt((x^3))
I wouldn't even call it a proof, but a way to see if both expressions denote the same function.
>>8060958
Pretty much.
if f(x) is the same as g(x), then the following equation must hold:
f(x)-g(x) = 0 => f(x)=g(x).
>>8061046
>No, you cannot do that I and would take points off of homework when students did that when I was grading for an intro to proof writing course.
Congratulations on being bad at math I guess?
>>8061136
He doesn't even know what proof by contradiction is. Why are people like that allowed to grade homework?
ITT I prove x + 2 = x + 1.
Start by writing x + 2 = x + 1
(x + 2) * 0 = (x + 1) * 0
0 = 0
Therefore x + 2 = x + 1, hurp durp
>>8061183
No, dumb idiot.
We assume that x+2 = x+1
We then do algebraic manipulation, and arrive at:
x = x-1
But this is a contradiction.
Therefore, it is not the case that x+2 = x+1
Ever heard of proof by contradiction, dumb angsty high school shitposter?
>>8061183
And if you multiply both sides by 0, you eliminate the x, and it isn't an expression of a function anymore, dumb cunt.
You can derive correct conclusions from wrong assumptions.
>>8061222
Exactly. This is the way proof by contradiction works.
>>8061210
>Ever heard of proof by contradiction, dumb angsty high school shitposter?
That's exactly a proof by contradiction.
Quote OP:
>>8060942
Let's say I need to prove that the expression of a function f(x) can be re-written in some other way than its original expression.
>To do that, can I just assume and state that both expressions of the function are equal, and then see how it goes from there? So that if I will arrive at an equality with same expressions on each hand, the proof will be completed.
Answer: No.
Counterexample: Take f(x) = x+2.
And just for you >>8061214
take g(x) = -x-2.
Obviously these two are not equal. However there exists a derivation that begins by assuming both expressions are equal and ends up with the same expression on both sides.
x + 2 = - x - 2
(x + 2)^2 = (-x - 2)^2
x^2 + 4x+ 4 = x^2 + 4x +4
According to OP, we "arrive at an equality with same expressions on each hand, the proof will be completed".
But clearly the proof is faulty since x + 2 != - x - 2.
Therefore OP's method is not allowed.
>>8061141
>He doesn't even know what proof by contradiction is. Why are people like that allowed to grade homework?
>implying he has ever graded homework
Remember: Any lies at all are evidence that the rest is an edifice built upon rot and corruption. He's a 13 year old kid playing hard on the internet.
>>8061243
f(x) = x+3
g(x) = -x-2
If f(x) = g(x), then for all real numbers,
x+3 = -x-2
We proceed to check if this is the case:
x+5 = -x
(-1)(x+5) = x
-x-5 = x
Contradiction.
Therefore, f(x) is not an equivalent to g(x).
It is a contradiction because clearly x is not equal to -x-5, for all real numbers. The OP was right. He simply forgot to add "for all real numbers".
>>8061268
That's not really a proof by contradiction. You show that f=g implies that x = -x-5 for all real numbers and apply the contrapositive. The obsession with and haphazard application of "proof by contradiction" in this thread is honestly the sign of a bunch of people who took a proofs class last semester and feel like experts now.
OP was not right. He suggests that if we assume two functions are equal and don't derive any untrue equalities then it must be true, and that is simply absurd as has been pointed out numerous times. Is it the same person going around this thread calling every correct post incorrect?
>>8061280
>(If the opposite is true, though, it would prove that the function can't be re-written as that expression)
Anon who he replied to here.
In fairness, OP goes on to state that
>(If the opposite is true, though, it would prove that the function can't be re-written as that expression)
and so far all the simple examples in this thread rely on using a faulty derivation to prove a false statement, but the catch with being simple is that there's always an alternate derivation that leads to an 'obvious' contradiction like 0 = 1 or whatever. (Which I think is axiomatically false in Peano arithmetic at least.)
The real problem with OP's statement is that it's not possible, without some sort of higher-order shenanigans, to prove that there are NO untrue derivations.
(Furthermore, unless I've misunderstood Godel's incompleteness theorem -- and plenty of people have -- it's possible to formulate two algebraic expressions that cannot be proved to be equal, via the usual trick of Godel numbering the derivation rules that I can't hope to fit in a single 4chan post, much less explain to the rest.)
>>8061268
This is not proof by contradiction you retards. You're just simplifying the expressions so you're brainlet minds can actually see the contradiction. What's stopping me from doing this:
x+3=-x-2
Contradiction
without showing any work? It's obvious they are not the same function from the very beginning, using algebra to get one side to be x is just mathematical masturbation.
Now you could use algebra to solve for x to find out where the functions intersect, and if x happens to cancel out and you get a statement like 2=2 then you can use this as a proof for them being the same function.
>>8061298
His exact phrase in the OP was
>So that if I will arrive at an equality with same expressions on each hand, the proof will be completed.
which is completely wrong. It shows a complete lack of understanding of the logic behind proving something so simple.
>>8061280
>>8061305
The OP was right, because if you arrive at an expression such that both sides would be IDENTICAL via algebraic manipulation, it would be the case that f(x) = g(x), for all x belonging to reals. It's a necessary condition that follows from algebra.
There's no proof by contradiction to be made. It's just basic algebraic manipulation.
ITT: Dumb brainlets not understanding basic algebra, nor proofs.
>>8061333
And here's why it's true:
f(x) - g(x) = 0 means that the expressions are identical
f(x) = g(x)
>>8061305
>which is completely wrong. It shows a complete lack of understanding of the logic behind proving something so simple.
It isn't even a proof in a strict sense - it's basic algebraic manipulation. But he technically isn't wrong. If he gets to the point where both sides are identical when setting f(x) = g(x), via algebraic manipulation, it will indeed show that f(x) is the same function as g(x).
Sadly, most /sci/ shitposters are dumb highschoolers who smoke too much weed and failed algebra.
>>8061354
>>8061333
Take f and -f. Suppose they're equal and square both sides. Since (f)^2 = (-f)^2, they must be equal? No, you absolutely cannot start by assuming what you want to show. If your chain of manipulations was sufficiently reversible, then you can actually put in the effort to construct a legitimate proof from it, but the OP statement was completely wrong.
>>8061359
My fucking God, are you dense?
(x)^2 = (-x)^2
x = sqrt(-x)
x = i*sqrt(x)
Contradiction. Why? Because you're implicitly involving imaginary numbers here.
This is some high school algebra here. Why is /sci/ divided over a dumb high school algebra question?
>>8061381
But I arrived at an equality! OP said if that if I assume f=-f and arrive at an equality then "proof will be completed." Can nobody fucking read?
>>8061398
No. I'm suggesting that those functions are equal for all reals, but have a different notation.
I'm also suggesting that the fact that they're equal for all reals but have a different notation doesn't solve the argument presented in this thread, because this function is an exception: it also involves imaginary numbers.
Hence, see:
>>8061381
For those who forgot - the argument is as follows:
Given two functions: f(x) and g(x), can one determine that f(x) and g(x) are identical functions by setting f(x) = g(x), and, via algebraic manipulation, arriving to the point where both sides of the equation are identical expressions? (Not just equal for all reals, but completely identical on both sides, symbol by symbol).
>>8061398
These functions aren't identical, but they're equal for all reals.
![acd[1].jpg](https://i.imgur.com/S2SILcA.jpg "acd[1].jpg")
The great proof pepe has awoken.
Some people in this thread clearly need to learn some basic proof theory.
What OP is trying to prove is f(x) <=> g(x).
To prove this, he needs to prove f(x) => g(x) and g(x) => f(x). Hence, he will need to derive g(x) from f(x) and f(x) from g(x), for this to be considered a worthwhile proof.
>>8061546
Here's a corrected example, based on a post made in this thread:
f(x) = sqrt(x^3)
g(x) = x^(3/2)
Prove that f(x) and g(x) are equivalent.
Proof:
We need to prove f(x) <=> g(x)
1) f(x) => g(x)
sqrt(x^3)
(x^3)^(1/2)
x^(3*(1/2))
x^(3/2)
2) g(x) => f(x)
x^(3/2)
x^(3*(1/2))
(x^3)^(1/2)
sqrt(x^3)
3) Since f(x) => g(x) and g(x) => f(x), we conclude that f(x) <=> g(x), and hence those two functions are equivalent.
>>8061092
>But why can't you assume that it's true and then attempt to derive a contradiction? If no contradiction is derived, doesn't it mean that the statement is correct?
No.
Multiply both sides by zero and you'll always get a true statement, but that doesn't mean your assumption is true.
This fallacy is called "begging the question".
>>8061260
Congratulations on realizing most of sci is retarded.
when I saw the OP i knew people where going say you can do it. Im just glad theres at least 2 other people who know you cant on sci.
