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This joke originally went in the "What cool little thing have you discovered today" thread. The joke itself was slightly off topic, and then through the magic of the internet the thread was derailed by math and formal logic. Apparently, being the first to make a math joke means that I now own the math thread. Brilliant.
Alright, I made an algebra error. Sqrt(-x) should be i*sqrt(x). So that inner term is just i*sqrt(bleep), which is just imaginary. Squaring that is just right back to -bleep.
What have I discovered today? I'm still a math nerd, even though I'm ... More than 20 years out of practice? Sheesh.
Alright, I made an algebra error. Sqrt(-x) should be i*sqrt(x). So that inner term is just i*sqrt(bleep), which is just imaginary. Squaring that is just right back to -bleep.
What have I discovered today? I'm still a math nerd, even though I'm ... More than 20 years out of practice? Sheesh.
The square root of a negative number is complex (imaginary). Squaring turns it into a real (if negative) number.
Of course, I have an advantage, as I go about my daily life attempting to internalize vast amounts of information in several different, mostly unrelated categories.
I have discovered/created (I made it, but it may already exist) a proof that given an infinite set of non-self-referential (aka given {A, B, C...} A cannot reference B or C or otherwise be made from them) true/false values you can prove that at least one of those values is false, in any possible situation, this can then be used to construct a lot of interesting things, logic is fun.
Consider an infinite set of true/false statements {A, B, C, D...}. To prove that at least one of them is false we must construct the following disjunction `(not A) or (not B) or (not C) or (not D)`. We can assume that given any X, X or (not X) is true. Therefore, given A we can assume A or (not A) is true. Then to prove our conclusion we must consider both scenarios, A or (not A). For the second scenario, we already have proven our conclusion, one of the values is false. For the first scenario, we invoke B or (not B), and then repeat the process infinitely. We then never run out of things that would cause the conclusion to be true, and thus it must be true as we cannot prove that given that A is true that B is true also and so on for all the set as they are not self-referential
The square root of a negative number is complex (imaginary). Squaring turns it into a real (if negative) number.
Of course, I have an advantage, as I go about my daily life attempting to internalize vast amounts of information in several different, mostly unrelated categories.
But if you're taking the sqrt of the number, the number itself is still real, so taking the sqrt and then squaring is no difference. However, if shit was defined as a variable that had some value that was complex then yes, it would "get real"
I call foul on your "proof", and invite you to consider the major change in logic after Godel demonstrated the flaws in logic proofs (that you are doing) with his unprovability theorems.
In a nutshell: You are in the same fallacy as proofs that assume "the set of all sets".
The infinite set of (true, ...) will not be self referential, and will not contain a false. Your proof will be (false or false or false or ...), and you are assuming that since you can keep going until you see a true, then there must eventually be a true.
This is not how proof by induction on integers work.
I have discovered/created (I made it, but it may already exist) a proof that given an infinite set of non-self-referential (aka given {A, B, C...} A cannot reference B or C or otherwise be made from them) true/false values you can prove that at least one of those values is false, in any possible situation, this can then be used to construct a lot of interesting things, logic is fun.
Consider an infinite set of true/false statements {A, B, C, D...}. To prove that at least one of them is false we must construct the following disjunction `(not A) or (not B) or (not C) or (not D)`. We can assume that given any X, X or (not X) is true. Therefore, given A we can assume A or (not A) is true. Then to prove our conclusion we must consider both scenarios, A or (not A). For the second scenario, we already have proven our conclusion, one of the values is false. For the first scenario, we invoke B or (not B), and then repeat the process infinitely. We then never run out of things that would cause the conclusion to be true, and thus it must be true as we cannot prove that given that A is true that B is true also and so on for all the set as they are not self-referential
I don't think I follow your proof. For one thing (this may be me being pedantic and you using slightly wrong terminology), a set cannot contain duplicate elements. So, if the elements are all either 'True' or 'False', then the biggest set you could possibly have is {'True', 'False'}. What I think you mean by "set" is a multiset, which can have duplicate elements.
Second, consider this infinite multiset:
S = {x > 0 | x ∈ N} = {1 > 0, 2 > 0, 3 > 0, 4 > 0...}
In the following proof, I will use the notation S[n] to refer to the nth element in the set when ordered as shown above.
Also, I will take as given that S[n] <=> n > 0.
I can prove by induction that all of the elements in this set are true.
Base case: S[1]
S[1] = 1 > 0 = true //This element is true, so the base case passes
Inductive step:
Assume n > 0 //We know this to be true
Prove n + 1 > 0 //so we want to use it to prove this true
n > 0 //Assume statement
n + 1 > 1 //Add 1 to both sides
1 > 0 //Do I really have to prove this?
n + 1 > 0 //Transitive property on previous two lines (think n + 1 > 1 > 0)
Q.E.D.
Due to the base case, we know that the first element in the multiset is true. The inductive step shows that, for any element, the next element is true. We can then apply it to the base case ad infinitum, proving that all elements in the set are true.
But if you're taking the sqrt of the number, the number itself is still real, so taking the sqrt and then squaring is no difference. However, if shit was defined as a variable that had some value that was complex then yes, it would "get real"
I have discovered/created (I made it, but it may already exist) a proof that given an infinite set of non-self-referential (aka given {A, B, C...} A cannot reference B or C or otherwise be made from them) true/false values you can prove that at least one of those values is false, in any possible situation, this can then be used to construct a lot of interesting things, logic is fun.
Consider an infinite set of true/false statements {A, B, C, D...}. To prove that at least one of them is false we must construct the following disjunction `(not A) or (not B) or (not C) or (not D)`. We can assume that given any X, X or (not X) is true. Therefore, given A we can assume A or (not A) is true. Then to prove our conclusion we must consider both scenarios, A or (not A). For the second scenario, we already have proven our conclusion, one of the values is false. For the first scenario, we invoke B or (not B), and then repeat the process infinitely. We then never run out of things that would cause the conclusion to be true, and thus it must be true as we cannot prove that given that A is true that B is true also and so on for all the set as they are not self-referential
But if you're taking the sqrt of the number, the number itself is still real, so taking the sqrt and then squaring is no difference. However, if shit was defined as a variable that had some value that was complex then yes, it would "get real"
I think you're taking the input data implications a little too far there. Either your variable is built around the fact that A is true or false, to put it in binary terms, A is either binary 1 or binary 0, or it is built around something that is completely irrelevant to whether A is true or false, in the latter case you haven't actually proven anything because you haven't tested A's variable state, you've only proven that you haven't actually tested it.
If A has to be a true or false variable, it must therefore be declared as such, and will have a value of either true or false. "A or (not A) is merely the result an examination of the specified variable in question, it still wouldn't examine A's actual variable state. To do that you have to construct a whole program that goes and seeds all the variables you want tested, to do that infintely requires infinite time, therefore your program will have entered a state of never being able to return from execution.
I don't think I follow your proof. For one thing (this may be me being pedantic and you using slightly wrong terminology), a set cannot contain duplicate elements. So, if the elements are all either 'True' or 'False', then the biggest set you could possibly have is {'True', 'False'}. What I think you mean by "set" is a multiset, which can have duplicate elements.
Second, consider this infinite multiset:
S = {x > 0 | x ∈ N} = {1 > 0, 2 > 0, 3 > 0, 4 > 0...}
In the following proof, I will use the notation S[n] to refer to the nth element in the set when ordered as shown above.
Also, I will take as given that S[n] <=> n > 0.
I can prove by induction that all of the elements in this set are true.
Base case: S[1]
S[1] = 1 > 0 = true //This element is true, so the base case passes
Inductive step:
Assume n > 0 //We know this to be true
Prove n + 1 > 0 //so we want to use it to prove this true
n > 0 //Assume statement
n + 1 > 1 //Add 1 to both sides
1 > 0 //Do I really have to prove this?
n + 1 > 0 //Transitive property on previous two lines (think n + 1 > 1 > 0)
Q.E.D.
Due to the base case, we know that the first element in the multiset is true. The inductive step shows that, for any element, the next element is true. We can then apply it to the base case ad infinitum, proving that all elements in the set are true.
I think the joke is that sqrt(-shit) would be imaginary shit. Squaring it gives real shit.
You can get the same punchline with complex shit if you multiply it by its complex conjugate.
I don't think you addressed the possibility that a variable in your data would be false there. All you did was prove that an imaginary multiset would all have true values, which would promptly be false if one of therm is actually false and therefore your multiset would promptly fold up like an accordion if it ever did encounter a false variable and therefore will never be heard from ever again!
For your multiset to be complete it must accept the possibility that there can be a false variable. And therefore, it must account for that possibility. If this was written as a computer program it'd have to have a false case scenario included, or it could never actually return from execution.
I call foul on your "proof", and invite you to consider the major change in logic after Godel demonstrated the flaws in logic proofs (that you are doing) with his unprovability theorems.
In a nutshell: You are in the same fallacy as proofs that assume "the set of all sets".
The infinite set of (true, ...) will not be self referential, and will not contain a false. Your proof will be (false or false or false or ...), and you are assuming that since you can keep going until you see a true, then there must eventually be a true.
This is not how proof by induction on integers work.
I'm not assuming that, I'm assuming you never reach a fail state of the proof (i.e. you can keep going), which is the same as saying that your conclusion must be true, and it is thus a fundamental flaw in logic itself (a contradiction inherent in the system, or at least the system given sets and infinities)
I don't think I follow your proof. For one thing (this may be me being pedantic and you using slightly wrong terminology), a set cannot contain duplicate elements. So, if the elements are all either 'True' or 'False', then the biggest set you could possibly have is {'True', 'False'}. What I think you mean by "set" is a multiset, which can have duplicate elements.
Second, consider this infinite multiset:
S = {x > 0 | x ∈ N} = {1 > 0, 2 > 0, 3 > 0, 4 > 0...}
In the following proof, I will use the notation S[n] to refer to the nth element in the set when ordered as shown above.
Also, I will take as given that S[n] <=> n > 0.
I can prove by induction that all of the elements in this set are true.
Base case: S[1]
S[1] = 1 > 0 = true //This element is true, so the base case passes
Inductive step:
Assume n > 0 //We know this to be true
Prove n + 1 > 0 //so we want to use it to prove this true
n > 0 //Assume statement
n + 1 > 1 //Add 1 to both sides
1 > 0 //Do I really have to prove this?
n + 1 > 0 //Transitive property on previous two lines (think n + 1 > 1 > 0)
Q.E.D.
Due to the base case, we know that the first element in the multiset is true. The inductive step shows that, for any element, the next element is true. We can then apply it to the base case ad infinitum, proving that all elements in the set are true.
This is self-referential, thereby by referencing n > 0 in your set in the induction you are not taking into account what I gave as the starting condition.
I think you're taking the input data implications a little too far there. Either your variable is built around the fact that A is true or false, to put it in binary terms, A is either binary 1 or binary 0, or it is built around something that is completely irrelevant to whether A is true or false, in the latter case you haven't actually proven anything because you haven't tested A's variable state, you've only proven that you haven't actually tested it.
If A has to be a true or false variable, it must therefore be declared as such, and will have a value of either true or false. "A or (not A) is merely the result an examination of the specified variable in question, it still wouldn't examine A's actual variable state. To do that you have to construct a whole program that goes and seeds all the variables you want tested, to do that infintely requires infinite time, therefore your program will have entered a state of never being able to return from execution.
Logic and programming are not the same at all . Also you don't have to examine A's actual state, you just need to prove that in both situations the conclusion is true.
I don't think you addressed the possibility that a variable in your data would be false there. All you did was prove that an imaginary multiset would all have true values, which would promptly be false if one of therm is actually false and therefore your multiset would promptly fold up like an accordion if it ever did encounter a false variable and therefore will never be heard from ever again!
For your multiset to be complete it must accept the possibility that there can be a false variable. And therefore, it must account for that possibility. If this was written as a computer program it'd have to have a false case scenario incled, or it could never actually return from execution.
What you have shown is that if I said it worked with self-referential sets (btw, when I say set in this context I mean multiset, or a set where the elements are statements that evaluateto but are not necessarily equal to true/false, that depends on definition) then you could prove that say 1 is not greater than 0, something that is obviously true (although, you could say self-referential truth statements as well and ould probs end with a fundamental logical flaw, yay)
I'm not assuming that, I'm assuming you never reach a fail state of the proof (i.e. you can keep going), which is the same as saying that your conclusion must be true, and it is thus a fundamental flaw in logic itself (a contradiction inherent in the system, or at least the system given sets and infinities)
This is self-referential, thereby by referencing n > 0 in your set in the induction you are not taking into account what I gave as the starting condition.
By that I mean the elements do not reference other elements in the set in any way shape or form, either in their proof of being true or in their statement
AKA, {A, B: proved by A, C: proved by B, D: proved by C...} is an example of a self-referential set
The self-referential thing is my own definition btw, I couldn't think of a good word for it
By that I mean the elements do not reference other elements in the set in any way shape or form, either in their proof of being true or in their statement
AKA, {A, B: proved by A, C: proved by B, D: proved by C...} is an example of a self-referential set
The self-referential thing is my own definition btw, I couldn't think of a good word for it
So, by "non-self-referential," you mean that you cannot determine the value of any given element even if you know the values of all the other elements? So, like, the elements are the results of random coinflips?
So, by "non-self-referential," you mean that you cannot determine the value of any given element even if you know the values of all the other elements? So, like, the elements are the results of random coinflips?
So... Take an irrational number, and express it in decimal. Examine each digit; if the digit is even, add a true to the multiset; if it's odd, add a false.
Now consider the number 0.24224222422224222224222222422222224222222224222222222422222222224...
So... Take an irrational number, and express it in decimal. Examine each digit; if the digit is even, add a true to the multiset; if it's odd, add a false.
Now consider the number 0.24224222422224222224222222422222224222222224222222222422222222224...
But the conclusion of your argument is that one element in any set that satisfies your (rather nebulous) requirement must be false. If you use that number (which is irrational) to fill in the elements of a multiset as described in the first line of the post you quoted, you'll find that all the elements are true. There can be no false elements because there are no odd digits in the number.
How, precisely, is that not a counterexample? (if it's not, tell me why- I'm trying to figure this out)
