How many equations would you say this expression represents, 1 2 or 3 or more?

 

[math] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} }[/math]

I would hazard at three. x,y and x,z and y,z -if you see what I mean

to clarify my thoughts - any less than three equations and information is lost and any more than three can be reduced to three

How many equations would you say this expression represents, 1 2 or 3 or more?

 

[math] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} }[/math]

 

How many equations would you say this expression represents, 1 2 or 3 or more?

 

[math] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} }[/math]

2

Formally, it is one line. I don't think that accounts for much. Syntactically, it is two equation signs. That is, you made two statements (note that "=" signs appearing in calculations also indicate statements, even if they are as simple as "I can re-arrange this term to look like that"). That accounts for a bit more. One may be tempted to arrange the three expressions in a triangle and write "=" signs between them to better show the symmetry. Accounts for not so much, because that is just my taste of aesthetics. Information-wise, it seems equivalent to two equations - the third equation gained by closing the equation circle does not contain any new information. That sounds like the most solid statement to me.

 

I wouldn't disagree with any number in {1,2,3}. If I was forced to pick one I'd go for 2.

The answer is certainly less than 3. Simply because you cannot get unique answers to the variables above, x, y, z. I base this on the old "if you have A unknowns, you need to have A equations to solve" rule. There are 3 variables above, but there isn't enough information to solve all 3, so I don't think that there are 3 equations.

 

I think that there are two.

 

x - A = y - B

 

and

 

y - B = z - C.

 

The fact that the 1st term and the 3rd term are also equal due to being equal to the middle term doesn't actually add any information.

Edited April 23, 201510 yr by Bignose

OK pedantry alert; the answer is certainly not "less than three".

The answer may be "fewer than three".

 

It's also possible that the answer is ill defined because there are (at least) two possible interpretations.

You can consider a=b as an equation or you can evaluate it as a calculation and the answer is either [true] or [false].

So if, for example a= 4 an b = 3 then a=b is false and (depending on a whole lot of arbitrary rules) that value may be coded as zero, or minus one (or possibly others).

So you could interpret the initial post as saying that the first part of the expression is zero if the second part is equal to the third and the first part is 1 if the second part is not equal to the third part.

 

In which case it's (arguably) not an equation at all.

 

Really, the point is that the post and question are ambiguous.

We have A=B=C. I think there is only two independent statement here (assuming '=' is what we usually mean etc) which we can write as

 

A=B and B=C.

 

As 'equals' is a symmetric relation we know B=A and C=B. (It is also reflective so A=A etc.)

 

'Equals' is also transitive and so if A=B and B=C then A =C. Which again being a reflective relation we also get C=A.

 

So I think it boils down to two statements and everything else follows.

How many equations would you say this expression represents, 1 2 or 3 or more?

 

[math] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} }[/math]

We can read the equations three types,

 

 

[latex] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} } [/latex]

 

 

[latex] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{z- z_{1} }{ z_{2}- z_{1} } [/latex]

 

 

[latex] \frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} } [/latex]

 

So, I think it would be 3 equations.

Edited October 15, 20169 yr by deesuwalka

'Equals' is also transitive and so if A=B and B=C then A =C. Which again being a reflective relation we also get C=A.

 

 

 

Another pedantry alert: I think the verb "equals" is a copula, and as such does not have a direct object. So it's not transitive.

 

Anyway, I read the OP as A = B and is also = C, so that is just 2 equations. A third equation adds nothing to it.

Edited October 15, 20169 yr by DrKrettin

It has 3 equations.

The first equation is the two point form of line equation in 2D coordinate system.

Edited October 15, 20169 yr by Sriman Dutta

Given A= B= C, we have immediately that A= B and that B= C. It follows from that that A= C. It is a matter of taste whether you think that last equation is "included" in the original rather than just "implied" by it.