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0

$begingroup$

I'm interested in equation/ inequality solving ( at the elementary, intermediate level: linear, quadratic, polynomial, absolute value, log, exponent, trig equations or inequalities ). My question is :

(1) what are exactly the legal operations when performing an equation/ inequality transformation

(2) in which case precisely the transformation produces an equivalent ( open) statement , so that one can put <=> between the two statements, in which case one is allowed to put an implication sign ( ==>) but not an equivalence sign?

algebra-precalculus logic

share|cite|improve this question

edited Mar 30 at 18:30

Mauro ALLEGRANZA

67.7k449117

asked Mar 29 at 16:37

Ray LittleRockRay LittleRock

10010

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, about “legal operations” on equations, if you remember that “$a=b$” means that $a$ and $b$ are not two numbers, but rather one and the same number, you really can’t go wrong. Anything you do to one side, you must do to the other, as long as you don’t try to divide both sides by zero (undefined operation) or take the square root of a negative number (undefined function).
  $endgroup$
  – Lubin
  Mar 29 at 17:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lubin. What about a < b?
  $endgroup$
  – William Elliot
  Mar 30 at 2:30
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm interested in equation/ inequality solving ( at the elementary, intermediate level: linear, quadratic, polynomial, absolute value, log, exponent, trig equations or inequalities ). My question is :

(1) what are exactly the legal operations when performing an equation/ inequality transformation

(2) in which case precisely the transformation produces an equivalent ( open) statement , so that one can put <=> between the two statements, in which case one is allowed to put an implication sign ( ==>) but not an equivalence sign?

algebra-precalculus logic

share|cite|improve this question

edited Mar 30 at 18:30

Mauro ALLEGRANZA

67.7k449117

asked Mar 29 at 16:37

Ray LittleRockRay LittleRock

10010

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, about “legal operations” on equations, if you remember that “$a=b$” means that $a$ and $b$ are not two numbers, but rather one and the same number, you really can’t go wrong. Anything you do to one side, you must do to the other, as long as you don’t try to divide both sides by zero (undefined operation) or take the square root of a negative number (undefined function).
  $endgroup$
  – Lubin
  Mar 29 at 17:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lubin. What about a < b?
  $endgroup$
  – William Elliot
  Mar 30 at 2:30
  

  
  
  
  

add a comment | 

$begingroup$

I'm interested in equation/ inequality solving ( at the elementary, intermediate level: linear, quadratic, polynomial, absolute value, log, exponent, trig equations or inequalities ). My question is :

(1) what are exactly the legal operations when performing an equation/ inequality transformation

(2) in which case precisely the transformation produces an equivalent ( open) statement , so that one can put <=> between the two statements, in which case one is allowed to put an implication sign ( ==>) but not an equivalence sign?

algebra-precalculus logic

share|cite|improve this question

edited Mar 30 at 18:30

Mauro ALLEGRANZA

67.7k449117

asked Mar 29 at 16:37

Ray LittleRockRay LittleRock

10010

$endgroup$

share|cite|improve this question

edited Mar 30 at 18:30

Mauro ALLEGRANZA

67.7k449117

asked Mar 29 at 16:37

Ray LittleRockRay LittleRock

10010

share|cite|improve this question

edited Mar 30 at 18:30

Mauro ALLEGRANZA

67.7k449117

asked Mar 29 at 16:37

Ray LittleRockRay LittleRock

10010

edited Mar 30 at 18:30

Mauro ALLEGRANZA

67.7k449117

edited Mar 30 at 18:30

Mauro ALLEGRANZA

67.7k449117

asked Mar 29 at 16:37

Ray LittleRockRay LittleRock

10010

asked Mar 29 at 16:37

Ray LittleRockRay LittleRock

10010

1 Answer
1

active

oldest

votes

1

$begingroup$

Algebraic "manipulations" on equations are based of the rules for equality.

For example, from $x=3$ we derive $x+2=5$ using the substitution axiom for functions :

$t = s → f[z/t] = f[z/s]$.

In the above case, we have to use the function $f(z) := (z+2)$.

In the case of an inequality, like e.g. $x<3$ we derive $x +2 < 5$ using the arithmetical theorem :

$a < b leftrightarrow (a + c < b + c)$,

that we can prove from Peano axioms.

The "simplifications", i.e. the manipulation leading from $x+2=3+2$ to $x+2=5$ in the first case, and from $x+2 < 3+2$ to $x+2 < 5$ are performed using the substitution axiom for formulas :

$t = s → (phi[z/t] → phi[z/s])$.

share|cite|improve this answer

answered Mar 30 at 18:30

Mauro ALLEGRANZAMauro ALLEGRANZA

67.7k449117

$endgroup$

add a comment | 

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1

$begingroup$

Algebraic "manipulations" on equations are based of the rules for equality.

For example, from $x=3$ we derive $x+2=5$ using the substitution axiom for functions :

$t = s → f[z/t] = f[z/s]$.

In the above case, we have to use the function $f(z) := (z+2)$.

In the case of an inequality, like e.g. $x<3$ we derive $x +2 < 5$ using the arithmetical theorem :

$a < b leftrightarrow (a + c < b + c)$,

that we can prove from Peano axioms.

The "simplifications", i.e. the manipulation leading from $x+2=3+2$ to $x+2=5$ in the first case, and from $x+2 < 3+2$ to $x+2 < 5$ are performed using the substitution axiom for formulas :

$t = s → (phi[z/t] → phi[z/s])$.

share|cite|improve this answer

answered Mar 30 at 18:30

Mauro ALLEGRANZAMauro ALLEGRANZA

67.7k449117

$endgroup$

add a comment | 

1

$begingroup$

Algebraic "manipulations" on equations are based of the rules for equality.

For example, from $x=3$ we derive $x+2=5$ using the substitution axiom for functions :

$t = s → f[z/t] = f[z/s]$.

In the above case, we have to use the function $f(z) := (z+2)$.

In the case of an inequality, like e.g. $x<3$ we derive $x +2 < 5$ using the arithmetical theorem :

$a < b leftrightarrow (a + c < b + c)$,

that we can prove from Peano axioms.

The "simplifications", i.e. the manipulation leading from $x+2=3+2$ to $x+2=5$ in the first case, and from $x+2 < 3+2$ to $x+2 < 5$ are performed using the substitution axiom for formulas :

$t = s → (phi[z/t] → phi[z/s])$.

share|cite|improve this answer

answered Mar 30 at 18:30

Mauro ALLEGRANZAMauro ALLEGRANZA

67.7k449117

$endgroup$

add a comment | 

$begingroup$

Algebraic "manipulations" on equations are based of the rules for equality.

For example, from $x=3$ we derive $x+2=5$ using the substitution axiom for functions :

$t = s → f[z/t] = f[z/s]$.

In the above case, we have to use the function $f(z) := (z+2)$.

In the case of an inequality, like e.g. $x<3$ we derive $x +2 < 5$ using the arithmetical theorem :

$a < b leftrightarrow (a + c < b + c)$,

that we can prove from Peano axioms.

The "simplifications", i.e. the manipulation leading from $x+2=3+2$ to $x+2=5$ in the first case, and from $x+2 < 3+2$ to $x+2 < 5$ are performed using the substitution axiom for formulas :

$t = s → (phi[z/t] → phi[z/s])$.

share|cite|improve this answer

answered Mar 30 at 18:30

Mauro ALLEGRANZAMauro ALLEGRANZA

67.7k449117

$endgroup$

share|cite|improve this answer

answered Mar 30 at 18:30

Mauro ALLEGRANZAMauro ALLEGRANZA

67.7k449117

answered Mar 30 at 18:30

Mauro ALLEGRANZAMauro ALLEGRANZA

67.7k449117

answered Mar 30 at 18:30

Mauro ALLEGRANZAMauro ALLEGRANZA

67.7k449117