Solution to Forbidden Equation: x=1 and 2

[calebhoilday]

I am not sure if the result I get for the equation below is forbidden or not.

((x+1)*(x-1))/(x-1)=3
x^2-1=3x-3
x^2-3x+2=0
(x-1)(x-2)=0

Their for x= 1 and 2

 

[tvavanasd]

Do both of your final answers work in the original equation?

 

[calebhoilday]

putting the value 1 into the equation results in 0/0=3. 0/0= undefined, the problem for me is the process of how it was solved seems mathematically sound.

 

[HallsofIvy]

Your first step was multiplying both sides by x- 1. If x- 1= 0 (that is, if x= 1) that is NOT "mathematically sound" because multiplying both sides of any equation, true or false, by 0 results in the true equation 0= 0.

 

[Sirium]

I suggest you whenever you multiplie with x or with, for exemple (x-1) to put x><0 (which means x can be 0) or in second exemple x-1><0; x><1 (which means x can't be 1). That way when u get answers you can check are they forbidden solutions very easy.
Also insted of >< you can writte crossed = which should look something like / on =.
Sry for my bad english and i hope i helpped

 

[Gib Z]

From the starting equation it is implicitly implied that x can not be 1 as division is meaningless otherwise.

 

[calebhoilday]

some time ago, i became uncomfortable with the graph this produces. I considered instead of a hole at 1 there should be a vertical line. this was based on the idea that you do not know 0/0= as it could be anything based on AxB=C their for A=C/B. Using other intersecting lines to to determine the co-ordinates, just as in the beginning post, their is a vertical line rather than a hole.

I just feel a vertical line is a better depiction.

 

[Gib Z]

...you do not know 0/0= as it could be anything ...

No, in fact its the opposite. The definitely nothing, its undefined. The division operation is meaningless when dividing by zero, and by writing the function like that there is an implicit assumption that x=1 is not in the domain of the function.

 

[calebhoilday]

Consider this. When taught in high school about gradients, we are taught that m=rise/run. depicted often is a vertical line for an undefined gradient. The formula for that line is y=(infinity/0)x. When x is equal to any number besides zero it does not occupy that domain as y=(infinity/0) and is undefined, but when x=0 y=(0/0) and depicted is a vertical line, defining it as occupying all values. Does this mean that this high school depiction is false?

 

[Office_Shredder]

y=(infinity/0)x is not the equation of a line, so that doesn't really make any sense

 

[HallsofIvy]

Consider this. When taught in high school about gradients, we are taught that m=rise/run. depicted often is a vertical line for an undefined gradient. The formula for that line is y=(infinity/0)x. When x is equal to any number besides zero it does not occupy that domain as y=(infinity/0) and is undefined, but when x=0 y=(0/0) and depicted is a vertical line, defining it as occupying all values. Does this mean that this high school depiction is false?

I cannot speak for all high schools- certainly I did NOT learn that in high school. But whether it is a "high school depiction" or not, yes, it is false!

 

[calebhoilday]

To my understanding the mathematical consensus is that any value divided by zero, is undefined. when I look at 0/0 compared to 1/0 i tend to find that they are different. Are they still both undefined, even if they are different?

Consider

X*0=1

There is no value as to which X could be that will have the result be 1 and hence if X = 1/0 then 1/0 is undefined.

Consider

X*0=0

X could assume any value at all and the equation holds true; hence if X=0/0 then my logic would consider that 0/0 is equal to any value.

How is this logic, bad logic?

 

[HallsofIvy]

Yes, 0/0 is "different" from a/0 for a non-zero. Some textbooks use the word "undetermined" for a limit, as x goes to a, that, if you just plugged a in for x would give 0/0.

But your logic is bad when you say "0/0 is equal to any value". The limit of something of the form "0/0" (if is exists) is some specific value- we just don't know which. But, in any case, 0/0 itself is not a number. The graph of y= (x^2- 2)/(x- 2) is <b>almost</b> a straight line. It is identitical to the graph of y= x+ 2 except at x= 2 where there is a hole. There is not a vertical line at x= 3 because 0/0 is NOT any number- certainly not &quot;all numbers&quot;.

 

[calebhoilday]

im not sure that I am getting you, but you say that 0/0 if it were some value would have to be a specific value. If (y-1)(y-2)=x then if X=0 and Y= 1 and 2, so to my understanding a particular domain can have multiple results. If you consider all those little tricks that can be played with proofs that consider 0/0=1 having 0/0=all values can give a better explanation. you often see the 1=2 proof, but if you were to say that 1*1=2*1 isn't correct, but 1*0/0=2*0/0 is correct as both would simply =0/0

Im sorry to push this perspective, its just i see it as being more advantageous to have 0/0 = all values, as it highlights that it is different X/0 (X not=0). (-1)^0.5 can't exist either but there is a whole branch of maths dealing with it.

Did you mean (X^2-4)/(x-2) ?

 

[Mark44]

Yes, I'm confident that HallsOfIvy meant (x^2 - 4)/(x - 2).

In your example, with (y-1)(y-2)=x, and x = 0, y can't simultaneously be 1 and 2.
If (y-1)(y-2)= 0, then EITHER y = 1 OR y = 2.

Your statement "so to my understanding a particular domain can have multiple results." makes no sense to me.

The trouble with an indeterminate expression of the form 0/0 is that if you could conceivably define it to be any value. For example, if you decided to define it to be 5, you would have 0/0 = 5. Now multiply both sides of the equation by 0 to get 0 = 0, a true statement. The same thing would happen if you decided to define 0/0 as -3. The problem is that a quotient should have a unique answer. For that reason it is NOT advantageous to have 0/0 being any arbitrary number.

In all of the 1 = 2 proofs, one of the steps typically involves division by zero, leading you to a contradictory result.

(-1)^(.5) doesn't exist in the real numbers, but it does exist in the complex numbers.

 

[calebhoilday]

how about any value rather than all values?

 

[Mark44]

how about any value rather than all values?

If an expression can take on any value, it can take on all values, so you're saying the same thing twice.

A division problem has to have a single, unique answer, or none at all. It can't have multiple answers.

 

[Mark44]

depicted often is a vertical line for an undefined gradient. The formula for that line is y=(infinity/0)x.

No it isn't. The formula for a vertical line is x = K, where K is some real number. The formula for a vertical line doesn't even involve y!

When x is equal to any number besides zero it does not occupy that domain as y=(infinity/0) and is undefined, but when x=0 y=(0/0) and depicted is a vertical line, defining it as occupying all values. Does this mean that this high school depiction is false?

 

[calebhoilday]

Im aware that high school depictions can be mislead. I really just think that this whole thing is more or less a matter of opinion.

To say a division problem, has to result in a single value, is more or less a rule someone has said. Show me the mathematical proof for this and ill back down from that statement.

I think that the idea of 0/0= any value should be played with to see if it has any application, because this is just a matter of opinion.

 

[jgens]

It's not a matter of opinion. An operation is defined in terms of functions, making it impossible for one input to produce multiple outputs. If you would like to define something differently, you can go ahead and do so, but you're probably not going to get very far.

 

[CRGreathouse]

The obvious definition caleb is herading toward is the preimage of [real] multiplication (here I'll use the symbol "/" for this, to remind you that it's division-inspired while not actually division), which is a "/" b = {a/b} for nonzero b, a "/" 0 = {} for nonzero a, and 0 "/" 0 = R, where R is the set of real numbers.

 

[jgens]

CRGreathouse, I'm not sure that that's the definition caleb is heading towards. Saying that 0"/"0 = R is considerably different than saying 0/0 = any element of R.

 

[Mark44]

Im aware that high school depictions can be mislead. I really just think that this whole thing is more or less a matter of opinion.

To say a division problem, has to result in a single value, is more or less a rule someone has said. Show me the mathematical proof for this and ill back down from that statement.

I think that the idea of 0/0= any value should be played with to see if it has any application, because this is just a matter of opinion.

How about addition? Is it a matter of opinion that the addition of two numbers should produce a single value? And multiplication? Is it also a matter of opinion that the product of two numbers should produce a single value? Division is an arithmetic operation the same as addition, subtraction, and multiplication, with the notable exception of not allowing division by zero, whether or not the dividend (the numerator) happens to be zero.

Arithmetic is so basic and has such a long history in humankind that there aren't a lot of books that present arithmetic at an axiomatic and theoretic level. The only one I'm aware of is Principia Mathematica, by Bertrand Russell and Alfred North Whitehead. I haven't read this, but I understand that they don't get around to proving that 1 + 1 = 2 until a ways into the second volume.

All four arithmetic operations that are defined produce a single value. The only arithmetic operation that is not defined is division by zero. Why would you want to make an exception for this particular operation and say that the result could be any number?

 

[CRGreathouse]

Arithmetic is so basic and has such a long history in humankind that there aren't a lot of books that present arithmetic at an axiomatic and theoretic level. The only one I'm aware of is Principia Mathematica, by Bertrand Russell and Alfred North Whitehead. I haven't read this, but I understand that they don't get around to proving that 1 + 1 = 2 until a ways into the second volume.

Metamath develops it; here's their 2 + 2 = 4, for example:
http://us.metamath.org/mpegif/2p2e4.html

The arithmetic above takes place in the complex numbers, and so the proof involves Dedekind cuts or the like underneath; not too simple. (I think complex numbers are pairs of reals, which are themselves Dedekind cuts of rationals, which are themselves ratios of integers, which are themselves pairs of natural numbers.) I seem to remember that there are simpler versions if you just want to add naturals.