1. ## Postulates behind algebra?

So what are the postulates/axioms that allow us to do the basic arithmetic operations to an equation, and then we come up with a new, but still equivalent equation with the first one? In practise: Why does an equation, which has been procecced using the same operations on both sides of the equation, have the same, yet only the same solutions as the original equation? And I want to tell you that I already know the "scale demonstration"/"balanced steelyard", which isn't the answer I'm lookin for. I need something more profound explanations.

2. Originally Posted by epson1
So what are the postulates/axioms that allow us to do the basic arithmetic operations to an equation, and then we come up with a new, but still equivalent equation with the first one? In practise: Why does an equation, which has been procecced using the same operations on both sides of the equation, have the same, yet only the same solutions as the original equation? And I want to tell you that I already know the "scale demonstration"/"balanced steelyard", which isn't the answer I'm lookin for. I need something more profound explanations.
the field axioms are what one typically thinks about fist when a question like this is asked because in most cases, the person asking the question has a limited knowledge of the different algebras for the different mathematical systems that are in existence today. So, to answer your question, namely
Originally Posted by epson1
So what are the postulates/axioms that allow us to do the basic arithmetic operations to an equation?

3. How is it exactly proven using the field axioms? I'm a high school student, so my understanding will be limited to what i have learned so far about maths( mostly one or two variable calculus, geometry, trig, algebra etc). Anyway I would be interested to see how the proofs go, and what kind of axioms are being used in them. This issue is also something that has brought me so much headache during my high school years.

4. So is it really so that nobody knows a proof to my question ? If so, I shall be very concerned about the truth value of our whole algebraic system.

5. Originally Posted by epson1
So what are the postulates/axioms that allow us to do the basic arithmetic operations to an equation, and then we come up with a new, but still equivalent equation with the first one? In practise: Why does an equation, which has been processed using the same operations on both sides of the equation, have the same, yet only the same solutions as the original equation? And I want to tell you that I already know the "scale demonstration"/"balanced steelyard", which isn't the answer I'm lookin for. I need something more profound explanations.
I think that the postulate you are looking for is the substitution property of equality, as defined here. But you should be careful about saying that "an equation, which has been processed using the same operations on both sides of the equation, [has] the same, yet only the same solutions as the original equation". That is not always true. For example, the equation $\displaystyle x=2$ has only one solution. But if you apply the operation of squaring to both sides of the equation then you get $\displaystyle x^2=4$, in which a new solution $\displaystyle x=-2$ has been introduced.

6. Originally Posted by epson1
How is it exactly proven using the field axioms?
It depends on what "it" is. If you give an example of a reasoning, we may try to show how it follows from the field axioms.

7. Originally Posted by emakarov
It depends on what "it" is. If you give an example of a reasoning, we may try to show how it follows from the field axioms.
Let's take the following example; 3x+1=2x-3. If we substract 2x from both sides, then we get x+1=-3. So why does these equations have the same, yet only the same solutions? We could continue with substracting 1 from both sides, giving us x=-4, which is the solution in this case. Now why did this procedure of adding 1 to both sides of the equation leave us with an equivalent equation with the previous equation?

I have also wondered another thing: If we have an equation that includes an unknown (possibly several unknowns), then how can we say that adding or multiplying both sides of the equation with the same number preserves the equality sign. For example, take the first equation: 3x+1=2x-3. Now we don't know yet which value of x satisfies the equation. If we now substract for example the number "1" from both sides of the equation, we would get 3x=2x-4. Notice that we still have the "=" sign between those two expressions, altough we don't even know if they are equal, because we don't know the correct number x. Or do we assume here that "x" is the correct solution to the equation, and then this presumption would allow us to preserve the "=" sign when substracting 1 from both sides? Tough if we did this presumption, then we would have a contradiction if the equation didn't have any solution at all.

8. Originally Posted by epson1
Let's take the following example; 3x+1=2x-3. If we subtract 2x from both sides, then we get x+1=-3. So why does these equations have the same, yet only the same solutions? We could continue with subtracting 1 from both sides, giving us x=-4, which is the solution in this case. Now why did this procedure of adding 1 to both sides of the equation leave us with an equivalent equation with the previous equation?
The answer is that these operations use the substitution property of equality that I mentioned in my previous comment (#5 above). This principle is absolutely basic in mathematics, and is really a postulate of mathematical logic rather than algebra. There is a very elementary and clear video showing the use of this principle at Watch Video on Substitution Property of Equality - Math Help on Yahoo! Video.

To see how that principle justifies the procedures used in solving equations, take the above example 3x+1=2x–3. Informally, we want to subtract 2x from both sides to conclude that x+1=–3. To justify that operation, first look at the left side of the equation. We know from rules of algebra (associativity, commutativity) that 3x+1–2x = x+1. The substitution property says that if 3x+1=2x–3 then we can substitute 2x-3 for 3x+1 in any equation, and the result will be the same. It follows that if (3x+1)–2x = x+1 then (2x-3)–2x = x+1. But 2x–3–2x = –3. Conclusion: x+1=–3.

The next operation (subtracting 1 from each side of that last equation) is justified in a similar way. We know that (x+1) – 1 = x. If x+1=–3 then the principle of substitution says that we can substitute –3 for x+1 and deduce that (–3) – 1 = x, in other words x=–4.

9. Originally Posted by epson1
I have also wondered another thing: If we have an equation that includes an unknown (possibly several unknowns), then how can we say that adding or multiplying both sides of the equation with the same number preserves the equality sign. For example, take the first equation: 3x+1=2x-3. Now we don't know yet which value of x satisfies the equation. If we now substract for example the number "1" from both sides of the equation, we would get 3x=2x-4. Notice that we still have the "=" sign between those two expressions, altough we don't even know if they are equal, because we don't know the correct number x. Or do we assume here that "x" is the correct solution to the equation, and then this presumption would allow us to preserve the "=" sign when substracting 1 from both sides? Tough if we did this presumption, then we would have a contradiction if the equation didn't have any solution at all.

How can you tell 3x+1=2x-3, before you have made sure x can be only -4? I mean you can't state that 3x+1=2x-3, unless you know that it really is true. For example we could say x+1=x, which is a false statement in the field of real numbers.

And thanks for introducing me the substitution property, because it is clearly the one of the axioms used in equations.

10. Originally Posted by epson1
...
If you can get your hands on Spivak's calculus, it explains what you're after in the first chapter.
(The book is really an introduction to mathematics in general just as much as it's one to calculus).

11. Originally Posted by TheCoffeeMachine
If you can get your hands on Spivak's calculus, it explains what you're after in the first chapter.
(The book is really an introduction to mathematics in general just as much as it's one to calculus).
Unfortunately I couldn't find that book in our local library, but anyway I can't still understand how the substitution property justifies the equivalence between some two equations.

12. Originally Posted by epson1
How can you tell 3x+1=2x-3, before you have made sure x can be only -4? I mean you can't state that 3x+1=2x-3, unless you know that it really is true. For example we could say x+1=x, which is a false statement in the field of real numbers.
You assume that there is some $\displaystyle a$ that when plugged into the polynomials $\displaystyle 3x+1$ and $\displaystyle 2x-3$ gives the same answer. Since you're assuming for now that there is such an $\displaystyle a$, you're allowed to manipulate things as if you have an equality. The result is to limit the values of $\displaystyle a$ that can work. Here we found that if there is such an $\displaystyle a$, then $\displaystyle a=-4$.

That still doesn't prove that there is a solution. Plugging $\displaystyle -4$ into the original equation proves that $\displaystyle -4$ is the only solution.

As for the equation $\displaystyle x+1=x$, assume there is a solution. Then we would have 1=0, which is false, so there are no solutions.

13. Originally Posted by LoblawsLawBlog
Plugging $\displaystyle -4$ into the original equation proves that $\displaystyle -4$ is the only solution.
Actually it only proves, that x=-4 is a solution, but not necessarily the only solution. How do you know there aren't any other solutions unless you have plugged every other number into the equation?

14. Originally Posted by epson1
Actually it only proves, that x=-4 is a solution, but not necessarily the only solution. How do you know there aren't any other solutions unless you have plugged every other number into the equation?
Well suppose that there are two solutions $\displaystyle x , y$ and $\displaystyle x \ne y$.

Then both

$\displaystyle 3x+1=2x-3$
and
$\displaystyle 3y+1=2y-3$

Must be true now subtract the 2nd equation from the first to get

$\displaystyle 3x-3y=2x-2y \iff x=y$ but this contradicts that there were two solutions. So if there is a solution it must be unique.

15. Originally Posted by TheEmptySet
Well suppose that there are two solutions $\displaystyle x , y$ and $\displaystyle x \ne y$.

Then both

$\displaystyle 3x+1=2x-3$
and
$\displaystyle 3y+1=2y-3$

Must be true now subtract the 2nd equation from the first to get

$\displaystyle 3x-3y=2x-2y \iff x=y$ but this contradicts that there were two solutions. So if there is a solution it must be unique.
Btw you used the assumption that the equivalence between for example the equations 3x-3y=2x-2y and x=y is valid. But that was exactly what was to be proven, that the equivalence between those two equations is true.

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