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I am attempting to learn linear algebra in a very thorough manner. The more questions I ask, the more I realize there are some fundamental issues I don't understand.

Here is one. For $x\in{}\mathbb{R}$, we learn to solve an equation like $5x=15$ by multiplying both sides of the equation by the multiplicative inverse of $5$, i.e., $\frac{1}{5}$. Someone tells us that this operation does not change the solution set of the equation, and most people will accept this. Thus we find that the solution set is $S=\{3\}$

However, multiplying both sides of the equation by 0 yields an equation $0=0$ with solution set $S=\mathbb{R}$.

Why the discrepancy?

Gary
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    The equation $0=0$ is still true so there is no discrepancy. We don't multiply by zero to solve an equation for exactly the reason you found: it doesn't help. – John Douma Dec 31 '21 at 17:32
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    See Why do we have to do the same things to both sides of an equation? AND In a group $G$ with operation $\star$, can I apply $\star$ to both sides of an equation? AND Losing Solutions to a Rational Equation. – Dave L. Renfro Dec 31 '21 at 17:40
  • @John Douma I mean a discrepancy in the solution set of the two equations. There is a discrepancy in that one equation has solution set {3} and the other has solution set R. – Gary Dec 31 '21 at 17:50
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    It's not a discrepancy. It's a loss of specifics. It would be analoguous to doing calculations to find Waldo is a 13 pound cat with black fur but a white left front paw, then doing a calculation and finding Waldo is an animal. It's not inconsistant but we lost information. And paw88789's answer explains why. Multiplying by 0 is not reversable. Another example of an irreversible is $x-1=5$ which has solution set ${6}$. But if we squared both sides (not reversable) we get $x^2-2x+1=25$ which has solution set ${6,-4}$. Not a descrepency; we lost specifics so we let more solutions slip in. – fleablood Dec 31 '21 at 18:08
  • @Gary You have erased both sides of the equation by multiplying by zero. This is equivalent to using Wite-Out to cover a word and its definition in your dictionary. Not only can't you find the word's definition, you can't even find the word. – John Douma Dec 31 '21 at 18:56
  • Have you tried multiplying both sides by the multiplicative inverse of $0$? – John Joy Jan 02 '22 at 15:20
  • @fleablood According to the American Heritage Dictionary, "discrepancy" means "An instance of divergence or disagreement. synonym: difference". There is a discrepancy between the solutions sets of the two systems. – Gary Jan 02 '22 at 15:35
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    @JohnJoy Joy The question is, why does multiplying by 0 change the solution set? Responding that the process is non-reversible/invertible doesn't seem to explain why. That would only explain why we can't guarantee that the solution sets would be the same, not explain why they are different. It seems maybe this is just something we must accept without understanding. – Gary Jan 02 '22 at 15:35
  • "There is a discrepancy between the solutions sets of the two systems. " No. THere is not. "$x$ is a real number" and "$x$ is $3$" are in complete agreement and there is no discrepancy at all. The only difference is on has more information and is more specific than the other. But as $3$ is a real number and "$x$ is a real number" means that $x$ is one of the real numbers and not "$x$ is all real numbers at the same time", there is utterly no discrepancy at all. – fleablood Jan 03 '22 at 06:32
  • @fleablood I appreciate your gusto but you're just using an different definition of the word "discrepancy" than I was. I see no disagreement on facts, only on word choice. – Gary Jan 03 '22 at 20:07
  • You asked why multiplying by zero makes a discrepancy in the solution sets. The simple fact is that it does not and there is no discrepancy. Every possible solution to $5x = 15$ is a solution to $0\cdot 5x = 0\cdot 15$. That $0\cdot 5x = 0\cdot 15$ has more solutions is not a discrepency. Multiplying both sides by $0$ allows in more solutions. Non-reversable operations can always allow in more solutions. But the never rule out solutions. Applying an operation to both sides never rules out solutions. That is not a discrepancy. – fleablood Jan 03 '22 at 21:48
  • The is no discrepancy (even with your definition) between "Waldo is an animal" and "Waldo is a 13 pound black cat with a white left front paw". There is no divergence and there is no disagreement. And "difference" can be used as a synonym for must cases but it is not an exact synonym for all instances. One can have a difference in specifics without being a discrepancy. – fleablood Jan 03 '22 at 21:54
  • @fleablood I'm really enjoying your vigor but at the same time I'm not convinced that "difference" isn't an exact synonym for "discrepancy". I'm old and that's my impression of a valid usage of the word, plus the dictionary agrees. It seems like you're taking the definition that means something like "contradiction". Again, we're talking semantics here, doesn't seem important. – Gary Jan 03 '22 at 22:43
  • But it is essential to your question. You ask why the solution set for $5x=15$ is ${3}$ but the solution set for $0\cdot x = 0\cdot 15$. You considered that somehow the solution set was changed and asked why the inconsistency. But ${3}\subset \mathbb R$ so there is no inconsistency. Multiplying by $0$ in non-reversable because there are multiple solutions to $0=0$ so all those multiple solutions are "let in". But it's not inconsistent as $5x=15$ is still maintained. – fleablood Jan 04 '22 at 01:35

2 Answers2

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Multiplying both sides of an equation by a nonzero number is reversible (invertible), and so information is preserved (i.e., the solution of the multiplied equation is the same as the solution of the original equation).

On the other hand, multiplying both sides by $0$ is not reversible, and so information is lost.

paw88789
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    This answer helped me in the following way. Hoffman and Kunze in their "Linear Algebra" (page 4) point out that if every equation in a system of linear equations (E2) is a linear combination of the equations in another system (E1), then all solutions of E1 also solve E2. However, there is no guarantee that the converse is true. As the equation 5x=15 is not a linear combination (i.e., scalar multiple) of the equation 0=0, there is no guarantee that the solution set will be the same. Thanks. – Gary Dec 31 '21 at 17:48
  • Mr. Paw, I do wonder about the logic of your statements though. Why is that that a non-reversible algebraic process must destroy information? Could there be cases in which a process is non-reversible but does not alter the original solution set? – Gary Dec 31 '21 at 17:57
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    It doesn't destroy information. It's just less restrictive in keeping potential solutions out. If we are doing something that isn't reversible we are allowing new potential answers (that are incorrect) in. [Note: The solution set of $A$ will ALWAYS be a subset of solution set of A with an non-reversible process applied. The solution set to $x-1=5$ will be a subset of the solutions set to $(x-1)^2 = 5^2$ which will be a subset of $\cos [(x-1)^2 = \cos(5^2)$ which will be a subset of the solution set to $0\times \cos [(x-1)^2 =0\times \cos(5^2)$] – fleablood Dec 31 '21 at 18:13
  • @Gary I suppose you could have something like the equation $0x=0$ (solution set is all real numbers). Multiply by $0$ and keep the same equation. – paw88789 Dec 31 '21 at 18:14
  • Certainly $1=2$ is false, but multiplying through by $0$ suddently makes "it" true. – Randall Dec 31 '21 at 18:55
  • @fleablood It certainly destroys information. The relation $5x=15$ contains more information than $0=0$ (the former basically tells us exactly what $x$ is, the latter tells us nothing). That information has thus been destroyed by multiplying with $0$. The very fact that we gain solutions when non-invertible operations are performed is exactly what we mean when we say we destroy information. Information is the quantity we need to narrow down the candidate solutions from the entire domain to our exact solution set, and destroying information corresponds exactly to losing capacity to narrow down. – Arthur Jan 01 '22 at 20:17
  • I will nitpick that the information isn'tt destroyed-- we merely chose to ignore it and throw it away. The information is still there and we needed to hold onto it. As for why non-reversible processes "destroy" (or as I'd say "ignore") information is because if it in non-reversible that means the path we took to get there is not the only one and we must no consider the potential of these other paths; paths we know are wrong but could have been used. Example $x+2=3$ means $x=1$ is the only path. But if you square both sides $(x+2)^2=3^2$ this allows the path $x+2=-3$. we lost specifics. – fleablood Jan 03 '22 at 06:42
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Each change on an equation has to be referenced to an axiom. Multiplying a real number by zero is not a bijective function, because zero has not a multiplicative inverse. What you have done is defining a function f(x) := 5x. You search an argument x where the function value is equal 15.

f(x)=15

By multiplying both sides of the equation with zero you have changed your function f to g(x):=0x. If you change your initial conditions, that is your function f, you can not expect that the solution set S stays the same.

oggiman
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