Verifying that all solutions to a system of eq. are found

[Mr Davis 97]

I am given the following problem: Find all ordered pairs (a, b) such that ##2a + b = 12## and ##ab = 3##. Given this system of equations, I simply use substitution, and then solve the quadratic ##2a^2 - 12a + 3 = 0##. Solving this, I obtain two ordered pairs: ##\displaystyle (\frac{6 + \sqrt{30}}{2},~6 - \sqrt{30})## and ##\displaystyle (\frac{6 - \sqrt{30}}{2},~6 + \sqrt{30})##. These are all of the ordered pairs that I found. The problem asks to find all of the ordered pairs that satisfy the system. How can I be absolutely sure (through some kind of informal logic) that only two such ordered pairs exist, and these two are those ordered pairs?

 

[fresh_42]

I am given the following problem: Find all ordered pairs (a, b) such that ##2a + b = 12## and ##ab = 3##. Given this system of equations, I simply use substitution, and then solve the quadratic ##2a^2 - 12a + 3 = 0##. Solving this, I obtain two ordered pairs: ##\displaystyle (\frac{6 + \sqrt{30}}{2},~6 - \sqrt{30})## and ##\displaystyle (\frac{6 - \sqrt{30}}{2},~6 + \sqrt{30})##. These are all of the ordered pairs that I found. The problem asks to find all of the ordered pairs that satisfy the system. How can I be absolutely sure (through some kind of informal logic) that only two such ordered pairs exist, and these two are those ordered pairs?

You can rule out ##ab=0##, so you may substitute ##b=\frac{3}{a}## and multiply the equation by ##a## without changing the solutions.
Now you see, that the equation is of order ##2##, i.e. there cannot be more than ##2## solutions.
To verify that your solutions are valid, simply insert them into the original equations and see whether they satisfy them.

You could also argue geometrically. A line (eq.1) intersects a hyperbola (eq.2) in at most ##2## points.

 

[Lucas SV]

Because you have deduced it. Assume ##(a,b)## is a par of real numbers satisfying your equations. this implies your quadratic equations. The quadratic equation only has two solutions. so the initial assumption implies ##a## is either of the two values you found. but as fresh pointed out, ##a## uniquely determines ##b##. So indeed there can only be two solutions.

 

[fresh_42]

Because you have deduced it.

Ok, in this small example it might be correct, if the calculation steps are carefully described.
It is not true in general, however. A deduction is ##x \in A ⇒ x \in B##. It does not guarantee ##x \in B ⇒ x \in A##.
Therefore all deduction steps will have to be equivalence relations. Your wording "because you deduced it" often leads to errors.
There is a fundamental difference between a necessary condition and a sufficient condition!

 

[Lucas SV]

Ok, in this small example it might be correct, if the calculation steps are carefully described.
It is not true in general, however. A deduction is ##x \in A ⇒ x \in B##. It does not guarantee ##x \in B ⇒ x \in A##.
Therefore all deduction steps will have to be equivalence relations. Your wording "because you deduced it" often leads to errors.
There is a fundamental difference between a necessary condition and a sufficient condition!

Yes this is all true, that is why I put an explanation later, although maybe i didn't make myself clear that i was filling the steps. If you want a set theoretic proof, you can take ##A'## be the solutions of the quadratic equation and ##A## to be pairs of real numbers which solve the first equation. Then show ##\phi:A'\rightarrow A##, ##\phi(a)=(a,3/a)## is a set isomorphism. So ##A## and ##A'## must have the same number of elements, or cardinality if you like. Since we know ##A'## has two elements, so has ##A##.

The details of the proof come in computing ##A'## and in showing that ##\phi## is a well defined function, and is both injective and surjective.