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what does this mean? (1 Viewer)
- Thread starter bos1234
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darkliight
I ponder, weak and weary
Agreed with Iruka that it's not very clear, but you could assume it meant the 'usual' addition and multiplication since it's talking about the real numbers.
There are four things that we need to check for each operation, for addition they are:
1. x + (y + z) = (x + y) + z, for all x, y, z in R.
2. there is an identity (lets call it 0) such that x + 0 = 0 + x = x, for all x in R.
3. for every x in R, there exists an x' in R such that x + x' = x' + x = 0 (we call this x' the inverse of x).
4. x + y = y + x for all x, y in R.
And (almost) exactly the same for multiplication ..
1. x * (y * z) = (x * y) * z, for all x, y, z in R.
2. there is an identity (lets call it 1) such that x * 1 = 1 * x = x, for all x in R.
3. for every x in R (WITHOUT 0), there exists an x' in R such that x * x' = x' * x = 1 (we call this x' the inverse of x).
4. x * y = y * x for all x, y in R.
Finally, we need to check that * distributes over +, ie just checking that x*(y+z) = (x*y) + (x*z) for every x, y, z in R.
Notice the additive identity is the '0' we're all familiar with (because x + 0 = 0 + x = x) and the multiplicative identity is the '1' we're all familiar with (because x*1 = 1*x = x). The additive inverse of x is (-x) (because x + (-x) = (-x) + x = 0) and the multiplicative inverse of x (for x not equal to 0 as we noted above) is 1/x (because x*(1/x) = (1/x)*x = 1).
The rest is easy to check since we're so familiar with the real numbers.
The point of doing this is that once we identify something as a field, then everything we prove that holds in one field, because it's a field, must hold in all the other fields because they're fields.
There are four things that we need to check for each operation, for addition they are:
1. x + (y + z) = (x + y) + z, for all x, y, z in R.
2. there is an identity (lets call it 0) such that x + 0 = 0 + x = x, for all x in R.
3. for every x in R, there exists an x' in R such that x + x' = x' + x = 0 (we call this x' the inverse of x).
4. x + y = y + x for all x, y in R.
And (almost) exactly the same for multiplication ..
1. x * (y * z) = (x * y) * z, for all x, y, z in R.
2. there is an identity (lets call it 1) such that x * 1 = 1 * x = x, for all x in R.
3. for every x in R (WITHOUT 0), there exists an x' in R such that x * x' = x' * x = 1 (we call this x' the inverse of x).
4. x * y = y * x for all x, y in R.
Finally, we need to check that * distributes over +, ie just checking that x*(y+z) = (x*y) + (x*z) for every x, y, z in R.
Notice the additive identity is the '0' we're all familiar with (because x + 0 = 0 + x = x) and the multiplicative identity is the '1' we're all familiar with (because x*1 = 1*x = x). The additive inverse of x is (-x) (because x + (-x) = (-x) + x = 0) and the multiplicative inverse of x (for x not equal to 0 as we noted above) is 1/x (because x*(1/x) = (1/x)*x = 1).
The rest is easy to check since we're so familiar with the real numbers.
The point of doing this is that once we identify something as a field, then everything we prove that holds in one field, because it's a field, must hold in all the other fields because they're fields.