An interesting proposition Please read

[philosophking]

I've disagreed with my professor on this point, because I still cannot see how I am wrong. Consider the following equation:

If x is an integer,

(x^3 + x)/(x^4 - 1) = x/(x^2-1)

I'm asked to either prove, or disprove this, and so I arrive at the following (man I wish I knew TEX or whatever it's called... lol)

[x(x^2+1)]/[(x^2-1)(x^2+1)]=

=[(x^2+1)/(x^2+1)]*[x/(x^2-1)]=x/(x^2-1) for all x in Z

Okay, now this is what I argued. I said that since (x^2+1)/(x^2+1) = 1 for every integer x, then the two equations equal each other. I mean, when you take any number, and multiply it by 1, you get that number back, right??---closure under scalar multiplication! I get really mad over this sometimes. The statement is pure logic to me.

My professor argues that since the equations on the left hand side and right hand side are undefined at x=1, they do not equal each other for every integer. But first of all, when did undefined <i>not</i> equal undefined? I just can't comprehend this. If 1*another number doesn't equal that number, I just don't know what will go on next.

Thanks so much for your consideration of my thoughts. It's likely that I'm just confused, and if so, a (nonlengthy) explanation would be much appreciated. Thank you!

 

[Janitor]

Would I be putting words in your mouth if I claimed that you are maintaining that

1 * x/(x^2-1)=x/(x^2-1) for all x in Z, including x = 1.

So that in the case of x=1, you are saying

1 * 1/0 = 1/0.

Your professor is saying that last line is not a valid equation, I take it. I am no expert on mathematical etiquette, but I can kind of see his point.

 

[philosophking]

Yes

That is exactly what I am saying. how can one times a number not equal that number? That's like saying 1*a != (does not equal) a ...ahh! do you see my frustration? hehehe

 

[Janitor]

I feel your pain [said in my best Bill Clinton voice]. Someone who is more up on the finer points of mathematical logic than I will surely weigh in on this.

 

[outandbeyond2004]

It's been more than 1 hour since Janitor's last post, so I tremblingly venture the following opinion:

Division by zero is never kosher, even when you get the result n/0 = n/0, or else you have to use the lazy 8 symbol. That aint a real number. If x is not 1, either side of the problem equation equate to a real number. But both sides become the lazy 8 for x = 1, instead. You may have seen a proof for 2 = 1 that used the lazy 8 in a subtle way.

You always have to watch out for such proofs.

Latex instructions are here: https://www.physicsforums.com/showthread.php?t=8997&highlight=latex

You could have entered "Latex" in the search function.

 

[chroot]

Well, 1/0 is indeterminate. If you approach 1/0 from the right, you end up with positive infinity. If you approach it from the left, you get negative infinity. In other words, 1/0 could mean either negative or positive infinity. Since it's ambiguous, we say it's indeterminate. Any equation that involves an indeterminate quantity is itself indeterminate. The equation 1 * (1/0) = (1/0) is not really an equation at all, and says nothing. Since 1/0 isn't really a number, it can't be manipulated like one.

- Warren

 

[matt grime]

The equation is undefined at x=1 so it is not a valid equation for all integers. If you're going to insist that one undefined qunatity always equals another undefined quantity you're going to go wrong very often. 1/x and 1/sinx are both not defined at zero but are not in any reasonable sense equal (they both tend to infinity as x tends to zero, but infinity isn't a number that you ought to do arithmetic with).

 

[philosophking]

Thank you all for your help. I can sort of see how 1/x does not equal 1/x for all x. But we're not talking about limits, as chroot said. we're talking about an actual equation, although i can sort of see how the two translate. matt grime: it's true that 1/x and 1/sinx do not equal each other, but x and sinx are different functions, and i can see that difference obviously. I am not "going wrong very often", because i can tell the difference between those two. But, when it's the same function a, and all you're doing is multiplying it by 1...ahhh anyone else have good ideas?

Thanks again, though, for all of your input :)

 

[ahrkron]

Click on the tex'd expressions to see how they are created:

Consider the following equation: If x is an integer,
\frac{x^3 + x}{x^4 - 1} = \frac{x}{x^2-1}

I'm asked to either prove, or disprove this, and so I arrive at the following
\frac{x(x^2 + 1)}{(x^2-1)(x^2+1)} = \frac{x^2+1}{x^2+1} \times \frac{x}{x^2+1} = \frac{x}{x^2+1} for all x in Z

Then you say that

since (x^2+1)/(x^2+1) = 1 for every integer x, then the two equations equal each other.

With "the two equations equal each other", I take it you mean "the two sides of the equation equal each other", but please correct me if not.

Maybe you can just phrase it like:

Which proves that \frac{x^3 + x}{x^4 - 1} = \frac{x}{x^2-1} for all x\neq 1, and both sides are indeterminate at x=1.

That way, you are including the fact that they both misbehave at x=1, and rigorous mathematicians will not complain for your use of "equals" between entities it is not defined to handle.

 

[matt grime]

You are the person who asked "when did undefined not equal undefined" and I gave you the starting point. If you want to learn more then look up relative and essential singularity, otherwise read, think and learn (sin(x)/x is, as x tends to 0?)

 

[Hurkyl]

Does blue fifths equal red squared?

 

[philosophking]

Thanks

Hurkyl, you make a good point, and matt thanks for your input. I will definitely look up those terms. I don't want to cause anymore confusion though, so I'll stop arguing this point that is obviously moronic and very unmathematical of me :(.