How do you distinguish between an identity and an equation?by tahayassen Tags: distinguish, equation, identity 

#19
Jan2513, 07:09 PM

P: 5,462

Hello, disregardthat, you have brought in the issue of 'completedness' in terms of qualifiers.
Equally I might observe that the statement [itex]\forall[/itex]x is incomplete. What x? [itex]\forall[/itex]x [itex]\in[/itex] R? [itex]\forall[/itex]x :→x>0? etc So where do you draw the line (stop)? 



#20
Jan2513, 07:15 PM

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disregardthat,
In post #15, I said 



#21
Jan2513, 10:34 PM

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These definitions are taken from the Oxford Concise Dictionary of Mathematics.
identity: An equation which states that two expressions are equal for all values of any variables that occur, such as x^{2}  y^{2} = (x + y)(x  y) and x(x  1)(x  2) = x^{3}  3x^{2} + 2x. Sometimes the symbol is used instead of = to indicate that a statement is an identity. equation: A statement that asserts that two mathematical expressions are equal in value. If this is true for all values of the variables involved then it is called an *identity, for example 3(x – 2) = 3x – 6, and where it is only true for some values it is called a *conditional equation; for example x^{2} – 2x –3 = 0 is only true when x = –1 or 3, which are known as the *roots of the equation. 



#22
Jan2613, 12:31 AM

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#23
Jan2613, 10:49 AM

P: 273

I always thought the ≡ symbol meant "is defined as". But I guess if it's defined as something, then it's also an identity. I think if they used that symbol more, then there would be less confusion for new students about the concept of an identity.
Thanks for the clearup. :) 



#24
Jan2613, 12:37 PM

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I've seen some texts that borrow notation from the Pascal programming language, :=, to indicate that a term or variable is being defined. 



#25
Jan2613, 01:07 PM

P: 273

Wait a second...
[tex](x3)(x+2)(x)\\ ={ (x }^{ 2 }x6)(x)\\ ={ x }^{ 3 }{ x }^{ 2 }6x[/tex] should really be... [tex](x3)(x+2)(x)\\ \equiv { (x }^{ 2 }x6)(x)\\ \equiv { x }^{ 3 }{ x }^{ 2 }6x[/tex] High school teachers and professors should really teach the [itex]\equiv[/itex] notation more. It makes things easier to understand by removing ambiguity. In programming, they make that distinction I believe. 



#26
Jan2613, 01:09 PM

P: 5,462

Pascal is not the only language to use = as the 'assignment operator' Tayahassen is studying computing. You then get such programming statements as x=x+1 Which makes sense in computing terms. 



#27
Jan2613, 01:14 PM

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Mark44 or any other authority, this has long time bugged me, though I live with it. Wouldn't you agree that most math e.g. textbooks writes a sort of slang here? That a lot of the 'equations' really should be identities?
As in any old theorem at random  say if you say If f(x) = g(x)h(x) then f'(x) = g'(x)h(x) + g(x)h'(x) you are not wanting to say you have this f, g, and h and then it may happen that for some particular values of x that first equation might be satisfied, in which case the second one is also for those particular values? Or reformulating in a single [g(x)h(x)]' = g'(x)h(x) + g(x)h'(x) should really be an identity, as maybe more than half the 'equations' in the average textbook should be? 



#28
Jan2613, 01:48 PM

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Cbased languages use = for assigment (only) and == for equality. Your example, Studiot, can be confusing to beginning programmers, until they understand that x = x + 1 is not saying that x and x + 1 are equal (an impossibility), but that the value of the expression x + 1 is being assigned to the variable x. A different example that comes to mind is the technique of partial fractions decomposition, in which you break down a rational expression that represents a product into the sum of two or more simpler rational expressions. For example, let's look at ##\frac{2}{x^2 + x} = \frac{2}{x(x + 1)}## The idea is that we want to write ##\frac{2}{x(x + 1)} ## as ##\frac{A}{x} + \frac{B}{x+1}## The equation we get is really an identity $$ \frac{2}{x(x + 1)} \equiv \frac{A}{x} + \frac{B}{x+1}$$ meaning that the two expressions on the left and right must be identically equal (barring two obvious values of x). If we multiply both sides by x(x + 1), we get ## 2 \equiv A(x + 1) + Bx##. Doing some rearranging, we get ## 2 \equiv (A + B)x + A##. Where students who are new to this technique have problems is recognizing that the only way two polynomials can be identically equal is when the coefficients of their respective terms are equal. The polynomial on the left is really 0x + 2, amd the one on the right is (A + B)x + A. With the recognition that this is an identity, it's easy to see that A + B = 0 and A = 2, hence B = 2. 



#29
Jan2713, 11:29 AM

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P: 3,173

Statements in mathematics take place within some context. Equations and identities are statements. Equations have "solution sets", which are the set of "things" that satisfy them within the given context. (i.e. their variables represent elements of some "universal set"). An identity is an equation who solution set is "everything", in given context.
For example, xy = yx is an identiy in the context of the set of real numbers, but not in the context of the set of 2 by 2 matrices of real numbers. 



#30
Jan2713, 03:51 PM

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#31
Jan2713, 08:41 PM

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As an identity: 2 ##\equiv## (A + B)x + A As an identity, the equation above has to be true for all values of x. In my previous work I solved for the constants A and B, and found them to be A = 2, B = 2. As an ordinary equation: 2 = (A + B)x + A If we interpret the above as an ordinary equation (not an identity), given values of A and B, we could solve for the value of x that makes the equation a true statement. The work would look like this. 2  A = (A + B)x $$\Rightarrow x = \frac{2  A}{A + B}$$ If someone tells us the values of the constants A and B, we can determine the value of x. 


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