Connectors between lines of working

[caroline1]

Hi,

please can someone help me understand the right kind of connectors I should use when presenting my work.

For example, if I wanted to write down the steps of my working for solving a quadratic I would write this:

9x^2-39x-30=0
3x^2-13x-10=0
(3x+2)(x-5)=0
3x+2=0 or x-5=0
x=-2/3 or x=5

However I am to put in connectors => and <=>

I don't understand when I use the => and when to use the <=>(i think that when the steps could flow backwards ways then I use <=>, and when it doesn't flow both ways I use =>

In which case:

9x^2-39x-30=0
<=> 3x^2-13x-10=0
<=> (3x+2)(x-5)=0
=> 3x+2=0 or x-5=0
=> x=-2/3 or x=5Thanks :-)

 

[Greg]

(i think that when the steps could flow backwards ways then I use <=>, and when it doesn't flow both ways I use =>

In which case:

9x^2-39x-30=0
<=> 3x^2-13x-10=0
<=> (3x+2)(x-5)=0
=> 3x+2=0 or x-5=0
=> x=-2/3 or x=5Thanks :-)

I agree. Your work looks good to me. :D

 

[Evgeny.Makarov]

Solving equation means finding the precise set of its solutions, not its superset or a subset. That is, solving $f(x)=0$ means finding the set $A$ such that
\[
\forall x\,(f(x)=0\iff x\in A).
\]
If you show a chain
\[
f(x)=0\iff P_1(x)\iff \dots\iff P_k(x)\implies P_{k+1}(x)\implies \dots\implies x=x_1\lor\dots\lor x=x_n
\]
where $\lor$ means "or" and $P_i(x)$ are some statements (such as $g(x)=0$ for some $g$), then you can only conclude
\[
\forall x\,(f(x)=0\implies x\in\{x_1,\dots,x_n\})
\]
i.e., that $\{x_1,\dots,x_n\}$ is a superset of the set of solutions. Among $x_i$ there may be some numbers that don't satisfy the original equation $f(x)=0$.

Why do you think that $(3x+2)(x-5)=0$ implies $3x+2=0\lor x-5=0$, but not the other way around?

 

[caroline1]

Solving equation means finding the precise set of its solutions, not its superset or a subset. That is, solving $f(x)=0$ means finding the set $A$ such that
\[
\forall x\,(f(x)=0\iff x\in A).
\]
If you show a chain
\[
f(x)=0\iff P_1(x)\iff \dots\iff P_k(x)\implies P_{k+1}(x)\implies \dots\implies x=x_1\lor\dots\lor x=x_n
\]
where $\lor$ means "or" and $P_i(x)$ are some statements (such as $g(x)=0$ for some $g$), then you can only conclude
\[
\forall x\,(f(x)=0\implies x\in\{x_1,\dots,x_n\})
\]
i.e., that $\{x_1,\dots,x_n\}$ is a superset of the set of solutions. Among $x_i$ there may be some numbers that don't satisfy the original equation $f(x)=0$.

Why do you think that $(3x+2)(x-5)=0$ implies $3x+2=0\lor x-5=0$, but not the other way around?

Thanks for your reply. I've no idea the answer to your question! but if this is the case then wouldn't it always be <=>? every time you had a new line of working it would be <=>. So what about the last line - would that be <=> as well?
Sorry, I didn't really follow the symbols! Its been many years since I studied them. Please could you rephrase in words? thanks

 

[Evgeny.Makarov]

I've no idea the answer to your question!

In general, $xy=0$ holds if and only if $x=0$ or $y=0$.

but if this is the case then wouldn't it always be <=>? every time you had a new line of working it would be <=>.

It depends on what you mean by "working". If I write a short story consisting of declarative sentences, it does not give me the right to say that questions have no use in our speech.

So what about the last line - would that be <=> as well?

Don't be afraid: if you think that $3x+2=0$ holds if and only if $x=-2/3$, just say so. Doubting things that are obvious is not helpful.

Sorry, I didn't really follow the symbols! Its been many years since I studied them. Please could you rephrase in words?

I used the following notations.
\[
\begin{array}{rl}
\forall x\,P(x): & \text{for all $x$, it is the case that $P(x)$ holds}\\
x\in A:&\text{$x$ is an element of the set $A$}\\
\{x_1,\dots,x_n\}:&\text{the set whose elements are $x_1,\dots,x_n$}\\
{\lor}:&\text{or}
\end{array}
\]

Sometimes it is appropriate to use $\implies$. For example, $\dfrac{f(x)}{g(x)}=c\implies f(x)=cg(x)$, but the converse implication does not hold. But, as I said, the goal of solving an equation is to find the precise set of solutions, so ideally all transitions in solving an equation should be equivalences ("if and only if"). In the case of a fraction,
\[
\dfrac{f(x)}{g(x)}=c\iff (f(x)=cg(x)\text{ and }g(x)=\ne0)
\]
In practice, you probably don't continue rewriting $f(x)=cg(x)$ adding "and $g(x)\ne0$" to every line. Instead, you add a note telling you after solving $f(x)=cg(x)$ to check that every solution satisfies $g(x)\ne0$.

Another example is
\[
f(x)g_1(x)=f(x)g_2(x)\Longleftarrow g_1(x)=g_2(x)
\]
but the direction $\implies$ does no hold. In this case,
\[
f(x)g_1(x)=f(x)g_2(x)\iff (g_1(x)=g_2(x)\text{ or }f(x)=0)
\]
Then you would probably start a chain of equivalent transformations of $g_1(x)=g_2(x)$ and another chain for $f(x)=0$ and in the end take the union of the sets of solutions. If you incorrectly ignore solving $f(x)=0$ and only concentrate on $g_1(x)=g_2(x)$, then you may not get all solutions of the original equation.

Still another example is
\[
\sqrt{f(x)}=g(x)\implies f(x)=(g(x))^2.
\]
Framing this as an equivalence requires
\[
\sqrt{f(x)}=g(x)\iff f(x)=(g(x))^2\text{ and }g(x)\ge0.
\]
In practice you probably stop the chain of equivalences for $\sqrt{f(x)}=g(x)$ and start another one for $f(x)=(g(x))^2$ but make a note that every solution of the latter equation has to be checked against $g(x)\ge0$.

 

[I like Serena]

Thanks for your reply. I've no idea the answer to your question! but if this is the case then wouldn't it always be <=>? every time you had a new line of working it would be <=>. So what about the last line - would that be <=> as well?
Sorry, I didn't really follow the symbols! Its been many years since I studied them. Please could you rephrase in words? thanks

Hi caroline!

In practice I recommend using only $\Rightarrow$.

The $\iff$ is actually applicable for every step that you've showed, but it's fraught with danger.
It requires to verify that the other flow of direction is also valid, and it's easy to overlook cases where that is not the case.
And usually it's not necessary to painstakingly keep track and verify that the opposite flow is also valid.

 

[Evgeny.Makarov]

And usually it's not necessary to painstakingly keep track and verify that the opposite flow is also valid.

How is using only $\implies$ sufficient in solving an equation?

 

[I like Serena]

How is using only $\implies$ sufficient in solving an equation?

The one thing extra we need, is the verification that the solutions that we find actually satisfy the original equation.
That closes the circle.
(And it's a step that is often neglected.)

 

[Evgeny.Makarov]

As I wrote in post #5, some transitions involve implication in the opposite direction. Ideally one should know for each transformation whether it adds or removes solutions and make notes along the way that say how to combine the sets of equation roots obtained in various branches of the solution.