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- Thread starter Larry
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- Jan 30, 2012

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- Jan 26, 2012

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Please give an example.

CB

- Feb 13, 2012

- 1,704

Even if a little 'advanced' for the pre-algebra forum, an 'example' could be the following. Let's consider the first order linear DE...Please give an example.

CB

$\displaystyle x\ y^{\ '}-y=0$ (1)

... the general solution of which is found to be $\displaystyle y=c\ x$. The 'standard' solving approach to (1) is to divide both terms by x under the condition that $\displaystyle x \ne 0$ so that the (1) becomes...

$\displaystyle y^{\ '}- \frac{y}{x}=0$ (2)

The problem is: what can You do if the interval of definition of the solution of (1) includes the point x=0?...

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

You solve it on the side of the singularity you are interested in (that is where the initial condition is specified). The equation could be solved on each side of the singularity, but a single initial condition will not suffice except under special conditions (essentially that \(\lim_{x\to 0}y/x\) exists from the side with the initial condition).Even if a little 'advanced' for the pre-algebra forum, an 'example' could be the following. Let's consider the first order linear DE...

$\displaystyle x\ y^{\ '}-y=0$ (1)

... the general solution of which is found to be $\displaystyle y=c\ x$. The 'standard' solving approach to (1) is to divide both terms by x under the condition that $\displaystyle x \ne 0$ so that the (1) becomes...

$\displaystyle y^{\ '}- \frac{y}{x}=0$ (2)

The problem is: what can You do if the interval of definition of the solution of (1) includes the point x=0?...

Kind regards

$\chi$ $\sigma$

CB

- Feb 13, 2012

- 1,704

That means that, if You have an initial condition in a point $x_{0}<0$, then the solution You find isn't valid for $x \ge 0$?... if yes of course it would be a severe limitation in many pratical situations...You solve it on the side of the singularity you are interested in (that is where the initial condition is specified). The equation could be solved on each side of the singularity, but a single initial condition will not suffice except under special conditions (essentially that \(\lim_{x\to 0}y/x\) exists from the side with the initial condition).

CB

Kind regards

$\chi$ $\sigma$

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Well maybe I should have considered asking in another group. But it was in high school algebra that I first asked the question and couldn't get an answer. Here is an example:Please give an example.

CB

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

When x = 1 the left side is 1/2, the right side is 1 -1 +1 -1............

This caused me in Algebra I to question the allow-ability of division. My teacher thought I was being a wise-*** and made me feel like a fool.

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- Feb 7, 2012

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Many mathematical operations are subject to some kind of condition or restriction. In the case of division, the only restriction is that division by zero is not possible. Division will never lead to absurdities if you take care not to attempt to divide by zero.Well maybe I should have considered asking in another group. But it was in high school algebra that I first asked the question and couldn't get an answer. Here is an example:

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

When x = 1 the left side is 1/2, the right side is 1 -1 +1 -1............

This caused me in Algebra I to question the allow-ability of division. My teacher thought I was being a wise-*** and made me feel like a fool.

Another situation where you have to take note of an important condition is when summing an infinite series. The binomial formula $1/(1+x) = x -x^2 +x^3-\ldots$ only holds subject to the restriction that $|x|<1.$ If you ignore that restriction and try putting $x=1$ then you get a result that has no mathematical basis. In this case, it is not division that is causing the apparent absurdity, but the essential condition required for the convergence of a series.

The lesson to learn from this is that many if not most mathematical statements come with some "small print" attached in the form of restrictions or conditions. You need to observe these conditions carefully if you want the results to make mathematical sense. If you ignore them, then you should not be surprised if it sometimes leads to absurdities. Having said that, I think that your teacher should have taken the opportunity to emphasise the importance of "reading the small print", rather than making you feel like a fool.

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- Jan 26, 2012

- 890

You cannot normally integrate through a singularity. Think about what happens when you integrate the ODE numerically away from the initial condition towards the singularity, you get to a point where the derivative is undefined (arbitrarily large absolute value) and you cannot step any further in that direction. On the other side of the singularity you could continue, but you now have no initial value to integrate away from.That means that, if You have an initial condition in a point $x_{0}<0$, then the solution You find isn't valid for $x \ge 0$?... if yes of course it would be a severe limitation in many pratical situations...

Kind regards

$\chi$ $\sigma$

(In practice you will either miss the singularity, but the result is nonsense beyond it, or you will hit it and get division by zero)

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Division by zero is not allowed because it leads to absurdities. Example, 4\0 = k, then k x 0 = 4, then 0 = 4. Absurd agreed. How come then sqrt{-1} which also seems absurd is assigned the symbol i and all is well?.....Many mathematical operations are subject to some kind of condition or restriction. In the case of division, the only restriction is that division by zero is not possible. Division will never lead to absurdities if you take care not to attempt to divide bynewline zero. The lesson to learn from this is that many if not most mathematical statements come with some "small print" attached in the form of restrictions or conditions....

I find it hard to accept that math is full of fine print. My gut still tells me if division, which is allowed, leads to absurdities, then it must be fundamentally flawed. All these are questions that must have at one time or another gone through the mind of any "thinking" beginning math student. They certainly ran through mine and the teachers were of no help whatsoever. I might have majored in math if I had had some of my early questions properly answered.

- Jan 26, 2012

- 890

You can do exactly the same sort of thing with division by zero as is done with the square root of minus 1. We introduce a new ideal element to the reals "nan" so that a/0=nan for any real a. Then we have some new rules for manipulating the ideal element "nan". There are a number of ways of doing this (one with two ideal elements nan and -nan, and another with one ideal element and where nan=-nan). Google for extended real numbers.Division by zero is not allowed because it leads to absurdities. Example, 4\0 = k, then k x 0 = 4, then 0 = 4. Absurd agreed. How come then sqrt{-1} which also seems absurd is assigned the symbol i and all is well?

I find it hard to accept that math is full of fine print. My gut still tells me if division, which is allowed, leads to absurdities, then it must be fundamentally flawed. All these are questions that must have at one time or another gone through the mind of any "thinking" beginning math student. They certainly ran through mine and the teachers were of no help whatsoever. I might have majored in math if I had had some of my early questions properly answered.

However the extended reals essentialy do nothing that not allowing division by zero does for us.

There is a reason why you should work with what is conventional (at least until you complete a couple of years of university maths), that is we have many hundreds of years work with the system as it is and the problems we do have do not lie with division by zero.

Before complaining about division by zero being an exception to some rules of arithmetic maybe you should wonder why you have no problem with real numbers.

CB

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