Solving for Identity of x*y = x + 2y - xy

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The discussion revolves around finding the identity of the operation defined by x*y = x + 2y - xy. It is determined that there is no identity element for this operation since while e = 0 satisfies x*e = x, it fails for e*x. The conversation then shifts to another operation, x*y = |x+y|, where it is initially suggested that both 0 and -2x could be identities. However, it is clarified that -2x cannot be an identity as it depends on x, and ultimately, the absolute value function does not yield a consistent identity. The conclusion emphasizes that most operations do not possess an identity, although it is possible to adjoin one to a set.
IKonquer
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I'm having trouble finding the identity of an operation. Could someone check my work?

I'm trying to find the identity of x*y = x + 2y - xy
In order to find the identity, I need to solve x*e = x for e

<br /> \begin{align*}<br /> x*e &amp;= x\\<br /> x + 2e - xe &amp;= x\\<br /> 2e - xe &amp;= 0\\<br /> e(2-x) &amp;= 0\\<br /> \end{align*}<br />
In order for e(2-x) = 0 to be true e = 0

Then I need to check if x*0 = x, which is true.
And then check if 0*x = x, which is NOT true, since 0*x = 2x.

So there is no identity and no inverse for the operation x*y = x + 2y - xy
 
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Looks ok, IKonquer! :smile:
 
Wow. Thanks for the quick response micromass! :smile:

What about this one?

x*y = |x+y|

To find the identity, x*e = x

x*e = |x + e| = x

The absolute value can be broken up into two cases:

1) x + e = x
e = 0

In order to be an identity x*e =x and e*x = x
x*0 = |x+0| = x
0*x = |0+x| = x

So 0 is an identity element.

---------------------------

2)x + e = -x
e = -2x

In order to be an identity x*e =x and e*x = x
x*(-2x) = |x+(-2x)| = x
(-2x)*x = |(-2x)+x| = x

So -2x is an identity element.

---------------------

So it seems like both 0 and -2x are both identities over the operation. Is this possible? And if two identities can occur, then every number under this operation has two identities.
 
IKonquer said:
Wow. Thanks for the quick response micromass! :smile:

What about this one?

x*y = |x+y|

To find the identity, x*e = x

x*e = |x + e| = x

The absolute value can be broken up into two cases:

1) x + e = x
e = 0

In order to be an identity x*e =x and e*x = x
x*0 = |x+0| = x
0*x = |0+x| = x

So 0 is an identity element.

---------------------------

2)x + e = -x
e = -2x

In order to be an identity x*e =x and e*x = x
x*(-2x) = |x+(-2x)| = x
(-2x)*x = |(-2x)+x| = x

So -2x is an identity element.

---------------------

So it seems like both 0 and -2x are both identities over the operation. Is this possible? And if two identities can occur, then every number under this operation has two identities.

First of all, -2x can never be an identity. Your identity cannot be dependent on x!
Second, you made a mistake here:

In order to be an identity x*e =x and e*x = x
x*0 = |x+0| = x
0*x = |0+x| = x

|x+0| is not x, is it? :frown:
 
micromass said:
First of all, -2x can never be an identity. Your identity cannot be dependent on x!
Second, you made a mistake here:



|x+0| is not x, is it? :frown:

I see, |x+0| = |x| = +x or -x. As a result there should be no identity.

So is it usually the case that most operations don't have an identity?
 
IKonquer said:
I see, |x+0| = |x| = +x or -x. As a result there should be no identity.

So is it usually the case that most operations don't have an identity?

Uuh, well, it depends with what you mean with "usually" :smile: Let's just say that having an identity provides quite a lot of structure on the set, so in that regard, it's quite rare for an operation to have an identity. However, you can always adjoin an identity to every set. For example

\{1,2,3,4,5,...\}

has no identity for the normal addition. But if we adjoin 0 to the set, then we do have an identity on the set.

Let's just say that the most operations that you'll ever meet will have an identity :smile:
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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