{a,-a,0} ∈ Z and inverse under division, plus notation Q

  • #1
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{a,-a,0} ∈ Z

For a set to have inverse under an operation, all elements must be able to be combined with another element of the set under that operation, after which the product of combination under said operation yields the identity element of that operation.

My question is, this is a strict definition, but the set of integers under division, does it have an inverse or not?
0/0=undefined. So, Zero is the element which does not fall under this defnition, indeed, 0/0 is undefined. But the definition must be strictly followed? there is also no element which 0 can be divided by to give one....

Second question about notation, { } are used to denote sets. But if we want to denote generalised elements like a, -a, and 0 and say they are elements of the set of integers, do we use brackets or exclude them?

a,-a,0 ∈ Z OR {a,-a,0} ∈ Z
 

Answers and Replies

  • #2
pwsnafu
Science Advisor
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We care about integers with multiplication having an inverse. That is: given ##x \in \mathbb{Z}## does there exist ##y \in \mathbb{Z}## such that ##xy=yx=1##.
The only integers with multiplicative inverses are 1 and -1.
You seem to be confusing multiplicative inverse with additive inverse (and it's trivial to see every integer has one of the latter).

The notation ##a, -a, 0 \in \mathbb{Z}## is correct. ##\{a, -a, 0\} \in \mathbb{Z}## means the set containing the elements a, -a and 0 is an element of ##\mathbb{Z}## which is not true.
 
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  • #3
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We care about integers with multiplication having an inverse. That is: given ##x \in \mathbb{Z}## does there exist ##y \in \mathbb{Z}## such that ##xy=yx=1##.
The only integers with multiplicative inverses are 1 and -1.
You seem to be confusing multiplicative inverse with additive inverse (and it's trivial to see every integer has one of the latter).

The notation ##a, -a, 0 \in \mathbb{Z}## is correct. ##\{a, -a, 0\} \in \mathbb{Z}## means the set containing the elements a, -a and 0 is an element of ##\mathbb{Z}## which is not true.

Thank you for your anser, the notation bit i got, but maybe I am being more dense than needed, the rest,I did not understand. Please bare with me...

I believe the multiplicative inverse for the number set of all integers does not exist, as 1/a is not an element of the integers. .............the integers do have the identity property under multiplication, as 1*a = a for all a except zero.

Z={-∞….-3,-2,-1,0,1,2,3…∞}


When we say multiplicative inverse we usually mean it for the set of rational or real numbers. as 1/a is an element of these two sets...

correct?
 
  • #4
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zero must be excluded if identity and inverse are to be fullfilled.

Under the set of natural numbers: Identity and inverse exists for all N under division, but not multiplication.

For the natural numbers, we find that the identity element under division is ONE. Any n divided by one, gives n.

Multiplication: a*1= a, 1*1=1. - this is the identity property of multiplication. Therefore, to get the identity element and find whether an inverse property exists for all N under multiplication we must multiply by 1/a. However, 1/a is not in the set of natural numbers.

Division= a/1 = a. This is the identity property of division. To get the inverse, we do a/a = 1 for all natural numbers.

My question was, the definition of identity and inverse states that all elements must be included, in inverse, all elements must be able to be combined with their fellow members of the number set in order to produce the identity element of that operation.

1,2,3 ..... 1/1 = 1, 2/1 = 2, 3/1 = 3. We see that the natural numbers have identity under division, with the identity element being one.

1,,2,3..... 1/1=1, 2/2= 1, 3/3=1
. We see that the natural numbers also have inverse under division, when any N is divided by itself, we get the number one, which is also the identity element under division.

But, when integers are considered....0/1 = 0. 0/0 = undefined. We have a problem of not all elements being applicable to the strict definitions, therefore I ask, does the number set of all integers have inverse under division? we see that if we divide all integers by themselves we get one, also the identity element of division.

0/1=0 0/0 =undefined. these are the two exceptions where the identity property of division of one does not work, and 0/0 is undefined.

Natural numbers had identity under multiplication with one, until you add 0 to the set, 0*1= 0 . My question is, the definition includes ALL elements, including zero. So how strict is it?
 
  • #5
362
96
a,-a,0 ∈ Z

The additive inverse property: a+ -a=0,0+0=0
If any integer a, is added to –a (or vice versa) then the result is always 0. 0 + 0 = 0 too.


The additive identity property: ∓a+0=∓a
If any integer±a is added to 0,the result is the same integer.


Thus,under the operation of addition,we can see two things:
1.Each integer has an additive inverse to which,if it is added to yields 0.
2. For the operation of addition, the identity element is also 0.
Therefore,we say integers have inverse property under addition

The multiplicative inverse:a* 1/a=1……1/a∉ Z
The set of integers does not have an inverse under mutliplication,as 1/a is not an integer,but a rational.

The subtractive inverse: a-a=0, (-a)- (-a)=-a+a=0
The identity element under subtraction is 0 as: ±a-0= ±a

Thus, integers have an inverse under subtraction.
Note: any number, positive or negative, subtracted with itself, will always yield 0

inverse under division: a/a=1.Each integer divided by itself yields one,except for the integer 0

a/0=undefined for all a including a=0.
Identity element of division:a/1=a.
Thus,we say the integers have inverse under division for all elements except zero.
 
  • #6
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Under the set of natural numbers: Identity and inverse exists for all N under division, but not multiplication.
Division as operation is tricky because it is not symmetric. If 1 is the identity you would expect this to work for both arguments, e. g. 1/5 = 5 which is not true. Better stick to addition and multiplication as operations.
My question was, the definition of identity and inverse states that all elements must be included
All elements in your set. You can choose the set to be "the integers without 0" for example.

Please stop using larger fonts and bold everywhere, that makes the post harder to read.
 
  • #7
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96
Division as operation is tricky because it is not symmetric. If 1 is the identity you would expect this to work for both arguments, e. g. 1/5 = 5 which is not true. Better stick to addition and multiplication as operations.All elements in your set. You can choose the set to be "the integers without 0" for example.

Please stop using larger fonts and bold everywhere, that makes the post harder to read.

ok, thanks. Sorry, usually this is how i make my notes because of my poor vision..i forget everyone has different preferences
 
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