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This is coming from my first course in undergraduate analysis, and it's confusing to me how to show that some operation is "well-defined". For example, my professor left as something for us to figure out on our own, not homework, to show ourselves that if $a,b,c$ and $d$ are integers, and ($b,d\not=0$) that $$\left[\left(\frac{a}{b}\right)\right]+\left[\left(\frac{c}{d}\right)\right]=\left[\left(\frac{ad+bc}{bd}\right)\right]$$ is well defined. He then made an example that said: If $\frac{a'}{b'}\sim\frac{a}{b}$ and $\frac{c'}{d'}\sim \frac{c}{d}$, then $\frac{a'c'}{b'd'}\sim \frac{ac}{bd}$.

Also above in the brackets are supposed to be $2$ classes of element's.

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    $\begingroup$ The term "well defined" can be thought of as a fancy way of saying "makes sense". For example, the naive form of adding fractions $a/b \oplus c/d = (a+c)/(b+d)$ doesn't make any sense, because if you change the way you write a fraction then the answer might change, e.g., $1/2 \oplus 1/3$ is $2/5$, but $2/4 \oplus 1/3 = 3/7$, and the answers are not the same fraction. So this operation doesn't make sense. The whole point is that when the way you compute an operation depends on how you write the inputs then you need to check different ways of writing inputs doesn't change the answer. $\endgroup$
    – KCd
    Jan 28, 2013 at 14:36
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    $\begingroup$ This stuff only arises when you define an operation on equivalence classes of things in terms of a representative from an equivalence class. As another example, the "fake integral formula" $I_{a,b}(f(x)) = F(b)+F(a)$, as opposed to $F(b) - F(a)$, where $F'(x) = f(x)$ on $[a,b]$. This makes no sense since it uses a choice of antiderivative, and if two antiderivatives differ by $C$, their sums at $a$ and $b$ differ by $2C$. Subtracting values of an antiderivative, on the other hand, kills the $C$ and thus is well-defined (that $F(b)-F(a)$ has significance in calculus is a separate matter). $\endgroup$
    – KCd
    Jan 28, 2013 at 14:40
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    $\begingroup$ Related link: gowers.wordpress.com/2009/06/08/… $\endgroup$
    – Micah
    Jan 28, 2013 at 16:19

4 Answers 4

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It usually means that the result doesn't depend on some more or less arbitrary choices you have to make.

One example is when you're dealing with equivalence relations (as in your case). The definition of a sum of two equivalence classes is made by choosing representatives for each class. To make this work well, you want the result to be independent of how you choose those representatives.

One other example is if you define a curve integral of a vector field using a parametrization of the curve. In that case you want to show that it doesn't matter which parametriztion you choose; you will still end up with the same value of the integral.

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To show that addition is well-defined there means to show that using different elements to represent the same equivalence class leads to the same result. You need to show that if $[\frac{a}{b}]=[\frac{a'}{b'}]$ and $[\frac{c}{d}]=[\frac{c'}{d'}]$, then $[\frac{ad+bc}{bd}]=[\frac{a'd'+b'c'}{b'd'}]$.

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    $\begingroup$ So what kind of proof would this require? Would I just need to define that elements are from the equivalence class and therefore we get the result in the last sentence? $\endgroup$ Jan 31, 2013 at 2:33
  • $\begingroup$ I think I'm getting what you need to know. So my professor, previously in the notes, said that if $\sim$ is an equivalence relation on S, {${b\in S|a\sim b}$} is $[a]_\sim$. Is this what you would need to know? $\endgroup$ Jan 31, 2013 at 2:52
  • $\begingroup$ $a\sim b$ means the pair $(a,b) \in \sim$. $\endgroup$ Jan 31, 2013 at 3:57
  • $\begingroup$ Ah yes that is it. My appologies, this is all very abstract and new to me. $\endgroup$ Jan 31, 2013 at 4:10
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    $\begingroup$ It's a shame that all the comments from the other party have been deleted. $\endgroup$ Oct 21, 2013 at 14:26
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In mathematics it is often the case that you have an interesting set $S$ of objects (the set ${\Bbb Q}$, the set of all curves of some sort, etc.), but the individual elements of this set do not have a unique presentation. Instead you have a much larger "set of representants" $\tilde S$ and a "production scheme" $\phi$ that for each representant $x\in\tilde S$ produces a unique element $\phi(x)\in S$.

Now you want to define (a) a property $P$ that some elements of $S$ might have and others don't, or (b) a function $f:\ S\to X$ from $S$ to some set $X$; but the only way to address the individual elements of $S$ is by means of their representants $x\in \tilde S$ and the scheme $\phi$.

The way to go about this is in case (a) that you define a property $\tilde P$ for the elements $x\in\tilde S$ that reflects your intentions. Then you prove that the element $\phi(x)\in S$ possesses the intended property $P$ iff $x\in\tilde S$ possesses property $\tilde P$. Thereafter you can say that property $P$ is well-defined on $S$.

Similarly in case (b): Here you define a function $\tilde f:\ \tilde S\to X$ and then have to prove that $\tilde f(x)=\tilde f(x')$ whenever $\phi(x)=\phi(x')$. Thereafter you can say that a "well-defined" function $f:\ S\to X$ has been established. It is related to $\tilde f$ and $\phi$ via $f\circ\phi=\tilde f$.

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Given two fractions $\frac{a}{b}$ and $\frac{c}{d}$, we need to show that the addition operation as defined by $\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$ is well defined.

As mentioned above, in order to do this, we need to show that if $\frac{a}{b}=\frac{a'}{b'}$ and $\frac{c}{d} =\frac{c'}{d'}$ then $\frac{a}{b}+\frac{c}{d}=\frac{a'}{b'}+\frac{c'}{d'}$

Which is equivalent to showing that $\frac{ad+bc}{bd}=\frac{a'd'+b'c'}{b'd'}$ or equivalent to showing that

[1] $(ad+bc)(b'd')=(a'd'+b'c')(bd)$

We begin with what we are given, that $\frac{a}{b}=\frac{a'}{b'}$ and $\frac{c}{d} =\frac{c'}{d'}$. This implies that

[2] $ab'=ba'$

[3] $cd'=dc'$

We multiply equation [2] by $dd'$ on both sides and multiply equation [3] by $bb'$on both sides and get:

[2] $(ab')(dd')=(ba')(dd')$

[3] $(cd')(bb')=(dc')(bb')$

We now add both equations and get:

$(ab')(dd')+(cd')(bb')=(ba')(dd')+(dc')(bb')$

We rearrange terms on the right side and the left side of the equation and get:

$(ad)(b'd')+(bc)(b'd') = (bd)(a'd')+(bd)(b'c')$

We pull out the common terms on each side and get:

$(ad+bc)(b'd') = (a'd'+b'c')(bd)$

This is equivalent to equation [1] which we wanted to show is true. This finishes our proof.

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