Fun fact: the Wikipedia page on psychological pain includes the sentence "Psychological pain is believed to be an inescapable aspect of human existence.[13]"
someone please see this as i am stuck for almost 1 day as whole . here i am writing in words , please see from my side . suppose you are police officer and there is an instruction written over wall that " if heart of a person stops then he is dead " . as a police officer you see this as an implication relation that is given between two statement " heart stops " and "person died" .
please let me write
Now according to his colleagues he was told that the statement can be false only when if a situation like " when the heart of person stops and he does not died " occurs . now you laugh as this can not be true
so whenever a case of murder came to you , you went to hospital and try to write statement of doctor
But you trust none but the instruction written on the wall
You catch a doctor telling lie when he says " the heart stops but he person does not died " beacuse you know this can not happen acoording to the instruction on the wall
now you go to doctor again this time he says " the heart does not stop but the person died " , now since the implication written on wall says that when the heart does not stop the doctor always tells truth
but now you come home ans see that doctor lied when saying " the heart does not stop but the person died" , but since the implication written on the wall allows doctor to say so you can not arrest him
because implication and biconditional have same values when given " p = true , q = false" and if for implication we are not concrned about when P= false hence we can never differentiate between both
i want to say that about implication " if heart stops then person dies " is false as it fails to adress the case when " p = false , q = true " .
also, I am fairly certain you meant that the same two statements cannot be connected by different ones (since that is what you had there). But neither is correct.
which operator will be used between the two statement "the heart stops " , " death occurs" . this " heart stops => death occur" or this " heart stops <=> death occurs "
@Zee please see some statement above i tried to tell about what actual case it , its a paragraph
@tobias , i think now we are on the same page . like why you only want to use P=true , Q= false as a decisive factor to deduce whether an implication is correct or not
here the contradiction occurs when "p=false , q= true " but since we have been told that in implication , we never bother about when P=false thus you drop this contradiction and try to state why even a illogical statement " heart stops and person not dead " could be true
why do not you use a case when P= false always makes an implication true , to contradict that in implication also P=false could lead to false statement ( a situation never occur , as this is how false is defined )
@TobiasKildetoft i do not mean that you do not know , but this mean i dont know what you know thus asking to tell not how to find answer but why to find answer in such a particular way
@AkivaWeinberger that is not the point , the point is the fact that " somebody living in utica could live in newyork state is possible and hence we say what we assumed for P= false , q= true in implication allign comes true allign with this possiblity
like " if you run you lose weight " is true implication ( by true i mean the relation between your running and loosing weight is implication )
exactly that is the point that we want to assingn a relation between statement but in case of implication its not always the case that when P=false, the whole implicaiton will come out to be true irrespective of the value of Q
because if it is so then we can never find a reason to differentiate " P=>Q " and " P <=> Q"
beacause in to check whether P=>Q is true you to check only a single case " P= true , Q= false " which by the also makes " P <=> Q " true
@Noob It seems that your issue is that you're saying, "If P => Q is true, then P = false Q = true is true, but in the heart implication that can't happen, so P => Q is false" but this is incorrect logic.
@Fagle but since other cases can happen , we made an assumption that when P = false we are entitled to belive that implicaiton is correct
Say a statement " if you have a dog then you can go inside the hall "
Now what this means that " If you do not have dog " then you do not need to read the full statement and whatever choice you make for going in or out of hall , the implication is true
i thing which is needed here is information . information about four possiblities
what four possiblities ? 1. what happened in case " a person had dog he went inside
2. " a person had a dog but did not went inside "
3. " a person did not had a dog but went inside "
4. " a person did not had a dog but did not went inside "
and if the value of cases 1,2,3,4, matches as " t, f, t , t" then implication is correct
else its not
this tell us that if the value of cases " 3 and 4 " does not came out to be true then the implication is false
but what happens in actual scenerio as YOU said " we do not care when P=false "
but as seen above that we do need the value of case 3,4 as true to determine whether the implication is true or not
so for statement " if heart stops then person dies " satisfies 1,2,4 but fails to satisfies 3 thus the implication is false because of " p = false " , q =true does not came out as true
which people do not care about as they do not find a possible case when the implication can be false when P=true
What you said is that "1 must be true, 2 must be false, 3 must be true, and 4 must be true" for an implication to be true. But 1 and 3, 1 and 4, and 3 and 4---none of those pairs can be true at the same time, so you have the definition wrong.
if 3, is false which occur when "P= false . q= true " but implication says when P= false then implicaiton is true thus contradiction occurs hence implication is false
what is understand is implication is relation between two statement , and because of our assumption to justify truth table as complete we said for P= false we take implication as true
If x = 3, then 2x = 6. And I'll also look the statement: x = 3 if and only if 2x = 6.
When x = 3, then "x = 3 and 2x = 6" (case 1) is true, so both statements are correct.
When x = 3, then "x = 3 and 2x =/= 6" (case 2) is false, so both statements are correct.
When x =/= 3, then "x =/= 3 and 2x = 6" (case 3) is false, so both statements are correct---or rather, "neither statement is incorrect".
When x =/= 3, then "x =/= 3 and 2x =/= 6" (case 4) is true, so both statements are correct.
In other words, cases 2 and 3 are always false, so either 1 or 4 is always true.
So P => Q and P <=> Q are both true.
The second one is more specific, but both are definitely true.
In general, case 3 is what you should check to see if they are different, but it might happen that case 3 is just impossible, in which case they're both correct.
To rephrase: P => Q is exactly the same as saying "case 2 never happens".
P <=> Q is exactly the same as saying "cases 2 and 3 never happen".
(Why? Because P <=> Q is the same as "P => Q and Q => P"----and do you see that Q => P's "case 2" is the original "case 3"?)
If case 2 never happens but case 3 does, then P => Q is right, and P <=> Q is wrong.
If cases 2 and 3 never happen, then they're both right.
So case 3 happening and case 2 never happening is the only way to tell them apart---but if case 3 never happens period, then they're both right or both wrong.
Once you pick a P (that doesn't vary) and a Q (that doesn't vary), P => Q and P <=> Q are fixed in stone. They're either true, or false. They might be different, but most often, they won't be.
So "which one is right" isn't really answerable in the nicest way. Either they're both right, neither of them are right, or the implication is right while the biconditional is wrong.
"If x = 3 then 2x = 6" isn't wrong, even though "x = 3 if and only if 2x = 6" is right. It's just less specific.
That's good. Personally, I think this is the kind of thing that can just give someone a headache if they think about it for too long. Implication really isn't the same as what we know of as implication in real life.
The truth tables are informed by intuition about what we'd like to be true mathematically.
Notice how the mathematical example of x = 3 iff 2x = 6 was a lot simpler than the heart example.
How to define smooth atlas to unit square; I guess the idea is we know that square is homeomorphic to circular disc and we can define smooth atlas there so somehow we need to use it
@Noob I also think that it might be difficult for you to grasp the if then logical statements if you don't speak English. Maybe that's one reason why you had problems understanding.
@jasper , it was never about grasping , it was more about knowing than mumbling definitions. apart from that , i am here to take notes of the discussion
@Noob It's good that you are trying to understand this basic logic, because this is the foundation of all mathematical reasoning.
@Noob Maybe we can summarise it as follows. "A implies B" is false if you can find an example where A holds but B does not hold. Otherwise, it is true. I think this makes it short and sweet.
@Noob Yes, but once you understand "A implies B", you will see that what I wrote pretty much sums it up in one line.
@Noob A test to check if you really understand something is this. After learning it, with all the proofs and symbols, try summing it up in a sentence or two, using words as much as possible. If you can do that, then good!
When we use words instead of symbols, we get to the core of our human language, which is the core of our human understanding.
For example, we can write y+y=2y. But if we say that something added to itself becomes twice of it, then that gets to the essence of the formula!
I am presently working through example 1.21 in Hatcher's book on wedge sums of topological spaces. He makes a few claims which I am having trouble verifying. First, let me set-up some notation.
Let $\{X_i\}_{i \in I}$ be a collection of topological spaces. Then $\amalg_{i \in I} X_i := \cup_{i ...
To show $S_3 \simeq D_3$, is it enough to say any symmetry of the equilateral triangle is a permutation of the vertices so there is (at least one) injective mapping $\phi : D_3 \to S_3$ and since $\mathrm{ord}(D_3) = \mathrm{ord}(S_3) = 6$ it's also (necessarily) bijective?
I haven't shown that $\phi$ is a homomorphism I guess
Suppose $G = \langle a,b \rangle$ and $H = \langle c,d \rangle$, that $|a| = |c|$ and $|c|=|d|$ (assume all orders are finite if necessary). Does it follow that $a \mapsto c$ and $b \mapsto d$ extends to an isomorphism of $G$ and $H$? I think the answer is no, but I can't think of any examples. I'm trying to find a $G$ abelian and $H$ non-abelian.
Hmm...Maybe it is true. Maybe I am being a knucklehead...
Ah, very nice. I think if both finite groups of the same order, then the theorem is true. The reason is that $a \mapsto c$ and $b \mapsto d$ will extend to a surjective*(?) homomorphism, and since they are finite of the same order, it will also be injective...Does this sound right?
What I'm trying to do is argue that $GL_2(\Bbb{F}_2) \simeq S_3$. Both are of the same order, and both are generated by two elements, one of order 2, the other of order 3. I think this is enough to conclude they are isomorphism, using the theorem (or some variant) mentioned above.
What I'm thinking is of taking two distinct $C_m\rtimes C_n$ (for suitable values of $n,m$), they are generated by the generators of the two cyclic groups, have the same number of elements and are not isomorphic
For an easy example one semidirect product can actually be direct so that one group will be abelian and the other won't
Here's a nice puzzle: Given $s$ ($s > 0$) points in the plane such that every three of them are contained in a disk of radius $1$. Prove that all $s$ points are contained in a disk of radius $1$.
Oh I figured out the answer to this
and it's so goddamn obvious
I'm disappointed in myself
(No idea if it's the solution I came up with the first time around)
I am presently working through example 1.21 in Hatcher's book on wedge sums of topological spaces. He makes a few claims which I am having trouble verifying. First, let me set-up some notation.
Let $\{X_i\}_{i \in I}$ be a collection of topological spaces. Then $\amalg_{i \in I} X_i := \cup_{i ...
