Logic

Ryan$ said:
THANKS ALOT VERY INFORMATIVE!
and yeah the teacher was reminding for of the formula, but is it allowed to write that?! that's my confusion ! I mean is it allowed to write I1=U/R ? and then =U1/R1 ??
Click to expand...
and may you please explain to me something I really struggle it every minute !!
I don't want to open new thread for it.

sometimes while solving a problem , I arrive to conclusion that to complete in my logic of the problem, then must *for instance* X be zero .. my confusing is, should I assign X to zero to complete consistent to my logic of the problem ?! if so then why?! maybe a senseable analogy to real life that's visualize my case why I should assign zero to complete my solution in logical way?! I'm talking in general case .. thanks alot
 
Ryan$ said:
THANKS ALOT VERY INFORMATIVE!
and yeah the teacher was reminding for of the formula, but is it allowed to write that?! that's my confusion ! I mean is it allowed to write I1=U/R ? and then =U1/R1 ??
Click to expand...
YES. Haven't I said that? What more do you need beyond my authority? ;)

Ryan$ said:
sometimes while solving a problem , I arrive to conclusion that to complete in my logic of the problem, then must *for instance* X be zero .. my confusing is, should I assign X to zero to complete consistent to my logic of the problem ?! if so then why?! maybe a senseable analogy to real life that's visualize my case why I should assign zero to complete my solution in logical way?! I'm talking in general case .. thanks alot
Click to expand...
As we've said before, we can answer you much better if you give us a specific example in which this occurs, including the actual problem statement and your work. I have no idea why you should do something in a problem you haven't shown.

Yes, you say it's a general question, but general questions include specific examples, which are often the best way both to understand the question fully, and then to explain it.
 
Does "a" appear in any part of this function outside the cosine? No.

So \(\displaystyle f(5) = sin^2 ( cos (5) )\)

End of story.

-Dan
 
You seem, in many of your posts, to think there is some "mystical" property to mathematics. Mathematical statements and definitions mean exactly what they say, nothing more, nothing less! If I define \(\displaystyle f(a)=sin^2(cos(a))\), where "a" can represent any number then anyone reading that knows immediately that \(\displaystyle f(5)=sin^2(cos(5))\), \(\displaystyle f(1000000)=sin^2(cos(1000000))\), \(\displaystyle f(x)=sin^2(cos(x))\), \(\displaystyle f(y)=sin^2(cos(y))\), \(\displaystyle f(t)=sin^2(cos(t))\), and \(\displaystyle f(*)=sin^2(cos(*))\), as long as x, y, t, and * represent numbers.
 
Hi guys!
well, there's something a lil confusing and hope we close this second gap, you could ask me for a question yeah I'll give you after I explain!
my problem is about "=" , I know what it's and what does it stand for but ! I stuck in a case when it's more complicated .. I mean like this:

if it's given that x=y and there's a formula (assume) x+y=3 then immediately it ring in my head that x=y then I assign instead of x, y in the formula ! (explicitly x,y given in the formula )

but second case is really confusing me !
if I have (M*H)/(x^2+y^2+z^2) then I know that's x=y .. but not assigning it into the formula because it has many parameters and can't decide "definitely" if the term after assigning x=y still the same or not ...and it's hard to ring to my head to assign or not in that case ............... so what should I do to overcome that problem? I mean to care about assigning and when I assign and is it allowed or not .. once again if it's given a formula in explicit form then yeah I assign immediately ...but in that case how really I could assign immediately? the formula is having many parameters and operators
 
Ryan$ said:
Hi guys!
well, there's something a lil confusing and hope we close this second gap, you could ask me for a question yeah I'll give you after I explain!
my problem is about "=" , I know what it's and what does it stand for but ! I stuck in a case when it's more complicated .. I mean like this:

if it's given that x=y and there's a formula (assume) x+y=3 then immediately it ring in my head that x=y then I assign instead of x, y in the formula ! (explicitly x,y given in the formula )
Click to expand...

Yes, once you know that x= y then you can replace "x" with "y" or vice versa. If you know both x= y and x+ y= 3 then you also know that x+ x= 2x= 3 and that y+ y= 2y= 3. So x= y= 3/2.

but second case is really confusing me !
if I have (M*H)/(x^2+y^2+z^2) then I know that's x=y
Click to expand...
 Wait, the way you have phrased that, with "then I know", it seems to be saying that because (MH)/(x^2+ y^2+ z^2) it follows that x= y. And that's not true! I assume you mean that you know (MH)/(x^2+ y^2+ z^2) and that, independently, x= y.

.. but not assigning it into the formula because it has many parameters and can't decide "definitely" if the term after assigning x=y still the same or not ...and it's hard to ring to my head to assign or not in that case ............... so what should I do to overcome that problem? I mean to care about assigning and when I assign and is it allowed or not .. once again if it's given a formula in explicit form then yeah I assign immediately ...but in that case how really I could assign immediately? the formula is having many parameters and operators
Click to expand...
Given that (MH)/(x^2+ y^2+ z^2) and x= y so that 2x^2 =2y^2 then both (MH)/(2x^2+ z^2)= (MH)/(2y^2+ z^2). All of the other parameters just "go along".
 
Hi guys, today I was on the school and I really confused so much about what Im going to ask!
the teacher wrote on the board:
x= 22/2+5
which afterward he said it's 16 !
but what is confusing me is who said that x is equal to the whol (22/2+5) without putting "(" into the term? I mean maybe x=22 without considering "/" and "+" .. who said that after equal we are dealing with the whole the term and not one of its objects?!

thanks in advance and I really confused every time that the right term after "=" is complicated I get confused but if it was like x=5 then it's fine for me ...not complicated ..
 
The whole left hand side of the equal sign equals the whole right hand side of the equal sign.

If this wasn't the case then x could equal 22 or 11 or 16 !
 
"but what is confusing me is who said that x is equal to the whol (22/2+5)."

The person who first defined "="! That is precisely what "=" means.
The purpose of parentheses is to make it clear that the quantity in the parentheses is to be treated as a single entity. There is no point in putting parentheses around the entire right side of an equation since it is already clear that it is a single entity by virtue of the fact that there is nothing else on that side of the equation.
 
Hi ! Im not asking who said that :) I want to learn why it's right and how could I think about it
lets assume given a=b=c , and in my problem I proved that c=d, then why certainly will also a=d , b=d ? what's confusing me is that Im not succeeding to imagine it .. I mean for instance I proved c=d and Im fine with this, and I know that a=b=c and Im fine with this, but!!! I didn't succeed to make the connection that also a=d ,b=d, so Im asking if there's any analogy is really describing "equality" that every time I face like this problem I remember it and could help me to understand the logic of equality ..
I mean for instance any analogy(a real life analogy would be really good) could be good approach to take it every time I see equal operator with more than one parameter ... thanks alot !!!
 
Equals is what's known as a binary equivalence relation. These have 3 properties.

a) Reflexive, i.e. for all x it's true that x = x

b) Symmetric, i.e. for all x, y it's true that x = y implies y = x

c) transitive, i.e. for all x, y, z, it's true that x = y and y = z implies x = z

using the transitive property you can show what you talk about in your post
 
3 + 4 = 7.

13 - 6 = 7.

4 * 2 - 1 = 7.

(50 - 1) / 7 = 7.

You REALLY cannot see that

3 + 4 = 13 - 6.

13 - 6 = 4 * 2 - 1.

4 * 2 - 1 = (50 - 1) / 7.

3 + 4 = (50 - 1) / 7.

EDIT: What romsek said can be viewed as expressing succinctly and axiomatically the observation that how we arrive at a specific numeric result is irrelevant.

This reverts to your questions a year or so ago when you asked why abstract generalizations were true without ever thinking about why they are true in specific cases.
 
Last edited:
Suppose a=b and a=c.
Then we have a-b=0 and a-c=0.
Subtract and get -b+c=0. So b=c.
There is the proof.
If you can't understand what JeffM you might not get this but this is the standard proof to show the transitive law of equality.

Just for the record, if you and I are the same age and you are also the same age as Jose you really don't see that that Jose and I are the same age?
 
Romsek said:
Equals is what's known as a binary equivalence relation. These have 3 properties.

a) Reflexive, i.e. for all x it's true that x = x

b) Symmetric, i.e. for all x, y it's true that x = y implies y = x

c) transitive, i.e. for all x, y, z, it's true that x = y and y = z implies x = z

using the transitive property you can show what you talk about in your post
Click to expand...
What you are saying is 100% correct, but the OP does NOT agree with your part c yet. This needs to be proven instead of just accepting. (there we go again with the difference between a math person and a brilliant engineer)
 
Jomo said:
What you are saying is 100% correct, but the OP does NOT agree with your part c yet. This needs to be proven instead of just accepting. (there we go again with the difference between a math person and a brilliant engineer)
Click to expand...
The transitive relation is not proven, it's taken as part of the definition so it doesn't have to be proven.

-Dan
 
topsquark said:
The transitive relation is not proven, it's taken as part of the definition so it doesn't have to be proven.

-Dan
Click to expand...
I think that you are mistaken. To have an equivalence relation you must satisfy the three conditions which Romsek noted. Why would we take the transitive property of equality as given when it can be proven.
Here is the proof. If a=b and b=c then a-b=0 and b-c=0. Adding yields a-c=0 or a=c.
It seems that you are saying that equality by definition is an equivalence relation, but that would be silly as many other binary operations are an equivalence relation. I think that equality is an equivalence relation since it follows the definition for an equivalence relation. What am I missing?
 
Jomo said:
What you are saying is 100% correct, but the OP does NOT agree with your part c yet. This needs to be proven instead of just accepting. (there we go again with the difference between a math person and a brilliant engineer)
Click to expand...
Ooops, you are the brilliant Physics, not the brilliant Engineer.
 
You must log in or register to reply here.
Top