Wikipedia:Reference desk/Archives/Mathematics/2017 February 1

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February 1

Logical Structure Behing Solving an Equation

Suppose I have an equation $f(x)=0$, and I know the solution exists and it is $x=a$. I perform a series of irreversible operations and arrive at the answer:

$f(x)=0\implies \cdots =\cdots \implies \cdots =\cdots \implies x=a$

which is just a big $f(x)=0\implies x=a$.

Since the operations are irreversible, we cannot go in the other direction and claim $x=a\implies f(x)=0$

Nevertheless, we know that the solution is $x=a$, so if we plug in $a$ into $f$ we get zero. But hold on a minute, the statement "if we plug in $a$ into $f$ we get zero" really means $x=a\implies f(x)=0$; we just went in the other direction, which is something we said we couldn't do!

What is wrong here? — Preceding unsigned comment added by 71.196.95.117 (talk) 01:34, 1 February 2017 (UTC)

For one thing, by plugging in $a$ you still don't know whether it is the only possible solution. ---Sluzzelin talk 01:40, 1 February 2017 (UTC)

Here, I presume that an "irreversible" implication is invalid in the reverse direction. Example: consider the function (all variables real) f(x) = (x − a)² + y², where y is an unknown and a is known. Then f(x) = 0 implies x = a, but the reverse implication does not hold. —Quondum 02:31, 1 February 2017 (UTC)

• "a is a solution to f(x)=0" is indeed logically equivalent to $x=a\implies f(x)=0$. If a is a solution, and you prove additionally that $f(x)=0\implies x=a$, then you have the full equivalence $f(x)=0\iff x=a$.

I suspect the confusion source lies somewhere in the "irreversible operations" part. They might not prove the reverse implication, but that implication may still be true for other reasons. Tigraan^{Click here to contact me} 10:44, 1 February 2017 (UTC)

The meaning of $f(x)=0\implies x=a$ is that a is the only possible solution. If you then found that f(a)=0 then you'd know a is the only solution, but if you found f(a)≠0 then you'd conclude there are no solutions. Not sure how the second case might come up in practice, but something similar could easily happen if you're finding, say, the minimum values of a function. --RDBury (talk) 07:07, 3 February 2017 (UTC)

Example:

$f(x)={\frac {x^{2}-2x+1}{x-1}}$

so

$f(x)=0\implies {\frac {x^{2}-2x+1}{x-1}}=0\implies x^{2}-2x+1=0\implies (x-1)^{2}=0\implies x=1$

but f(x) as written is not defined at x=1, so x=1 is not a solution to f(x)=0. Gandalf61 (talk) 10:24, 3 February 2017 (UTC)

Having trouble visualizing this solid

Bounded by z=y^2, x=0, and z=y-x. I know what each looks like but don't see how they all fit together. 69.22.242.15 (talk) 10:21, 1 February 2017 (UTC)

Never mind.

Resolved

— Preceding unsigned comment added by 69.22.242.15 (talk) 10:31, 1 February 2017 (UTC)