Problem with implication and bi-implication.

[ehj]

I have a problem which is best explained with an example:

sqrt(A)=B => A=B^2 <=> B=-sqrt(A) or B=sqrt(A)

Since A=B^2 <=> B=-sqrt(A) or B=sqrt(A) I should be able to reverse the order of this into

B=-sqrt(A) or B=sqrt(A) <=> A=B^2

But if this is done in the first equation you obviously run into a problem:

sqrt(A)=B => B=-sqrt(A) or B=sqrt(A)

What is it that I'm doing wrong, what wasn't allowed in above calculation?

 

[Kurret]

"sqrt(A)=B => B=-sqrt(A) or B=sqrt(A)". Notice the "or" there, if sqrt(A)=B then obviously the statement "B=-sqrt(A) or B=sqrt(A)" holds, since only one of B=-sqrt(A) or B=sqrt(A) needs to be true, and B=sqrt(A) is true...

 

[HallsofIvy]

I have a problem which is best explained with an example:

sqrt(A)=B => A=B^2 <=> B=-sqrt(A) or B=sqrt(A)

Since A=B^2 <=> B=-sqrt(A) or B=sqrt(A) I should be able to reverse the order of this into

B=-sqrt(A) or B=sqrt(A) <=> A=B^2

But if this is done in the first equation you obviously run into a problem:

sqrt(A)=B => B=-sqrt(A) or B=sqrt(A)

What is it that I'm doing wrong, what wasn't allowed in above calculation?

I presume you mean "(B=-sqrt(A) or B=sqrt(A)) <=> A=B^2" and "sqrt(A)=B => (B=-sqrt(A) or B=sqrt(A))". I see nothing wrong with either of those.

 

[ehj]

okay great

 

[evagelos]

HalsofIve the sqroot of 9 is 3 and not 3 or -3 but how do you guarantee that in the definition of sqroot??

If you write sqroot(A)=B <===> B^2=A ,only.... then sqroot(9)=B <===> B^2=9 <==> (B=3v B =-3) AND hence sqroot(9)= 3 or -3

So what is the proper definition of the sqroot??

 

[mrandersdk]

look here:

http://en.wikipedia.org/wiki/Square_root

 

[HallsofIvy]

The proper definition of square root is, of course, "sqrt(a) (for a a non-negative real number) is the positive real number b such that b²= a".

Now, back to the implications. Using that definition,
"(B=-sqrt(A) or B=sqrt(A)) <=> A=B^2" is correct.

The statement "sqrt(A)=B => (B=-sqrt(A) or B=sqrt(A))" is also a correct statement but has nothing to do with the definition of sqrt(A). It is true simply because "A=> (A or B)" is a tautology.

In particular, I did NOT write, nor would I, "sqrt(A)=B <===> B^2=A". That is false.

 

[ehj]

So if I have an expression, for instance a speed squared equals something, but I'm only interested in writing the positive solution, it would only be mathematically correct to write:

v^2 = b <= v=sqrt(b)

Right?

 

[HallsofIvy]

If you are only interested in the positive solution, you don't need to write "$\pm$". Although I think it would be better to write v^2= b => v= sqrt(b) since the implication is going that way.

 

[ehj]

But you said earlier: "(B=-sqrt(A) or B=sqrt(A)) <=> A=B^2" is correct."
The statement "sqrt(A)=B => (B=-sqrt(A) or B=sqrt(A))" is also a correct statement"
Which gives you that sqrt(A)=B => A=B^2
If that is rearranged and B replaced with v and A with b you get:
v=sqrt(b) => v^2=b
which is the same as
v^2=b <= v=sqrt(b)
Which was what I asked about the correctness of earlier.
If what you said is also true, that v^2=b => v=sqrt(b) you have implications both ways and actually a biimplication: v^2=b <=> v=sqrt(b) which is not correct! So it has to be either.. But you seem to be giving two different answers HallsofIvy :P

 

[HallsofIvy]

But you said earlier: "(B=-sqrt(A) or B=sqrt(A)) <=> A=B^2" is correct."
The statement "sqrt(A)=B => (B=-sqrt(A) or B=sqrt(A))" is also a correct statement"
Which gives you that sqrt(A)=B => A=B^2
If that is rearranged and B replaced with v and A with b you get:
v=sqrt(b) => v^2=b
which is the same as
v^2=b <= v=sqrt(b)
Which was what I asked about the correctness of earlier.
If what you said is also true, that v^2=b => v=sqrt(b) you have implications both ways and actually a biimplication: v^2=b <=> v=sqrt(b) which is not correct! So it has to be either.. But you seem to be giving two different answers HallsofIvy :P

No, it is the difference between "or" and "and".

If "A is true" then "either A is true or B is true" is a tautology.

"sqrt(A)= B=> (B= -sqrt(A) and B= sqrt(A))" is a false statement: "sqrt(A)= B" does NOT give "B= -sqrt(A)".

But "sqrt(A)= B=> (B= -sqrt(A) or B= sqrt(A)" is a true statement because "sqrt(A)= B" does give "B= sqrt(A)" and so the "or" statement is true whether "B= -sqrt(A)" is true or false.

I did not and would not say "v²= b <=> v= sqrt(b)".

I would say "v²= b <=> (v= sqrt(b) or v= -sqrt(b))"

 

[ehj]

"I did not and would not say "v2= b <=> v= sqrt(b)"." I know, I didn't say you did. I completely agree with your latest post, but I don't see it's relevance? I never mentioned "and" in any of the math?
The only problem I have now is that you said that v^2=b => v=sqrt(b)
It gives rise to a problem.
Do you agree that v=sqrt(b) => v^2=b ?
This has to be correct since I'm simply squarring both sides, applying the _function_ f(x) = x^2 on both sides. By the definition of a function, it relates only one element in the codomain for each element in the domain, which makes it impossible for the second statement v^2=b to be false. So v=sqrt(b) => v^2=b has to be correct. This is the same as v^2=b <= v=sqrt(b)
You wrote in an earlier post that v^2=b => v=sqrt(b). Clearly one of these has to be wrong since we would otherwise have a biimplication between the two statements (which is wrong, like you said yourself). You agree that A => B and B => A can be put together to A <=> B right? This post seems to be very much like the last i posted :/

 

[Moo Of Doom]

I think in post #9, Halls meant to switch the order of the statements, so that we don't have this awkward <= in the middle, but forgot to actually do it.

 

[HallsofIvy]

"I did not and would not say "v2= b <=> v= sqrt(b)"." I know, I didn't say you did. I completely agree with your latest post, but I don't see it's relevance? I never mentioned "and" in any of the math?
The only problem I have now is that you said that v^2=b => v=sqrt(b)
It gives rise to a problem.
Do you agree that v=sqrt(b) => v^2=b ?
This has to be correct since I'm simply squarring both sides, applying the _function_ f(x) = x^2 on both sides. By the definition of a function, it relates only one element in the codomain for each element in the domain, which makes it impossible for the second statement v^2=b to be false. So v=sqrt(b) => v^2=b has to be correct. This is the same as v^2=b <= v=sqrt(b)
You wrote in an earlier post that v^2=b => v=sqrt(b). Clearly one of these has to be wrong since we would otherwise have a biimplication between the two statements (which is wrong, like you said yourself). You agree that A => B and B => A can be put together to A <=> B right? This post seems to be very much like the last i posted :/

Please read ALL of post 9:

If you are only interested in the positive solution, you don't need to write "$\pm$". Although I think it would be better to write v^2= b => v= sqrt(b) since the implication is going that way.

I have added the emphasis.

This was in response to post 8:

So if I have an expression, for instance a speed squared equals something, but I'm only interested in writing the positive solution, it would only be mathematically correct to write:

v^2 = b <= v=sqrt(b)

Right?

where again I have added the emphasis.

 

[mrandersdk]

Please read ALL of post 9:

I have added the emphasis.

This was in response to post 8:

where again I have added the emphasis.

Maybe I'm reading you wrong, but assuming all is real numbers, it is always true that

v = sqrt(b) => v^2 = b

and

v = -sqrt(b) => v^2 = b

but

v^2 = b => v = +/- sqrt(b) (sqrt meaning the principle square root)

if you know that v is positive also then you of cause have

v^2 = b => v = sqrt(b).

But for it to mathematical correct it would be better to write

(v^2 = b and v > 0) => v = sqrt(b).

 

[ehj]

So you say that saying "v is positive" is like adding a condition like

v^2=b and v>0 => v=sqrt(b)

 

[mrandersdk]

yeah or else i would write

v^2 = b => v = +/- sqrt(b)

and then as a physicist, write something like: Because of the context (of some given exercise fx.), only the positive solution makes sense so the negative is rejected.

 

[ehj]

Maybe you can even say v^2=b and v>=0 <=> v=sqrt(b)
where >= means bigger than or equal to
?

 

[Tac-Tics]

yeah or else i would write

v^2 = b => v = +/- sqrt(b)

and then as a physicist, write something like: Because of the context (of some given exercise fx.), only the positive solution makes sense so the negative is rejected.

The plus-or-minus sign is an evil shorthand used to confuse students of algebra.

This is actually a case of abuse of the equals sign. When you see a teacher write that x^2 = 4 implies x = +/- 2, he or she means something that cannot be expressed using standard algebraic notation. He or she *means* to say that the *set of solutions* to that equation is {-2, 2}. He or she *means* to say that x = -2 OR x = 2. But it can't be both, and you cannot correctly express it using a single equals sign.

One thing that algebra teachers often neglect to teach their students is that the sqrt function is in fact NOT the inverse of squaring! The equation sqrt(x^2) = x is true sometimes, but not all the time. It's only true when x >= 0! Squaring is an operation that has no inverse. Once you square something, you can't undo it safely.