How do you distinguish between an identity and an equation?


by tahayassen
Tags: distinguish, equation, identity
Studiot
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#19
Jan25-13, 07:09 PM
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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)?
Mark44
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#20
Jan25-13, 07:15 PM
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disregardthat,

In post #15, I said
This is a good point, but might be a smidgen on the pedantic side. I would venture to say that most mathematics instructors would treat the qualifier "there exists an x such that..." as being implied. Logicians might quibble at this, but I doubt that mathematics instructors would.
Best Pokemon
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#21
Jan25-13, 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 x2 -
y2 = (x + y)(x - y) and x(x - 1)(x - 2) = x3 - 3x2 + 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 x2 – 2x
–3 = 0 is only true when x = –1 or 3, which are known as the
*roots of the equation.
Mark44
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#22
Jan26-13, 12:31 AM
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Quote Quote by Best Pokemon View Post
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 x2 -
y2 = (x + y)(x - y) and x(x - 1)(x - 2) = x3 - 3x2 + 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 x2 – 2x
–3 = 0 is only true when x = –1 or 3, which are known as the
*roots of the equation.
Thank you, Best Pokemon. These definitions look pretty close to what I have been saying in this thread.
tahayassen
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#23
Jan26-13, 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 clear-up. :)
Mark44
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#24
Jan26-13, 12:37 PM
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Quote Quote by tahayassen View Post
I always thought the ≡ symbol meant "is defined as".
No, the usual meaning is that the expressions on either side have the same value.

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


Quote Quote by tahayassen View Post
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.
Other texts distinguish between identities and conditional equations by using ##\equiv## vs. =.
tahayassen
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#25
Jan26-13, 01:07 PM
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Wait a second...

[tex](x-3)(x+2)(x)\\ ={ (x }^{ 2 }-x-6)(x)\\ ={ x }^{ 3 }-{ x }^{ 2 }-6x[/tex]

should really be...

[tex](x-3)(x+2)(x)\\ \equiv { (x }^{ 2 }-x-6)(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.
Studiot
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#26
Jan26-13, 01:09 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.
Good point.

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.
epenguin
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#27
Jan26-13, 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 equation expression relationsionship

[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?
Mark44
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#28
Jan26-13, 01:48 PM
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Quote Quote by Studiot View Post
Good point.

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.
Many programming languages distinguish between assignment and equality by using different symbols. In Pascal-based languages such as Modula-2 and (I think) Ada, the assignment operator is := and the equality operator is =.

C-based 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.

Quote Quote by epenguin View Post
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?
Yes on both.
Quote Quote by epenguin View Post
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?
These equations are essentially identities. We are defining f to be the product of g and h, pretty much independent of the value of x (as long as x is in the common domain of g and h, of course). The equation with f' can also be considered an identity.
Quote Quote by epenguin View Post

Or reformulating in a single equation expression relationsionship

[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?
I agree. It would probably cut down the confusion.

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.
Stephen Tashi
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#29
Jan27-13, 11:29 AM
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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.
tahayassen
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#30
Jan27-13, 03:51 PM
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Quote Quote by Mark44 View Post
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.
If ≡ was replaced with =, wouldn't it also be true that the coefficients of their respective terms are equal?
Mark44
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#31
Jan27-13, 08:41 PM
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Quote Quote by tahayassen View Post
If ≡ was replaced with =, wouldn't it also be true that the coefficients of their respective terms are equal?
No, because the two equations would have different meanings. Let's look at the equation in my example a couple of posts ago.

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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