# What are a b?f(x)=ax+b isomorphism in structure rings (r,+,*), (r,t,o) xty=x+y-2 xoy=1/4xy-1/2(x+y)+3

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### 1 Answer

You need to find the identity element for `x t y = x + y - 2` , hence, there exists an element `e_t in R` , such that for each element `x in R` , the following identity holds:

`x te_t = e_t t x = x`

`x +e_t - 2 = x => e_t = 2`

You need to find the identity element for `xoy = (1/4)xy - (1/2)(x+y)+3` , hence, there exists an element `e_o in R` , such that for each element `x in R` , the following identity holds:

`x o e_o = e_o o x`

`(1/4)xe_o - (1/2)(x + e_o) + 3 = x`

`(1/4)xe_o - (1/2)(x) - (1/2)(e_o) + 3 = x`

You need to isolate the terms that contain `e_o` to the left side, such that:

`(1/4)xe_o- (1/2)(e_o) = x - 3 + x/2`

You need to factor out `(1/2)(e_o)` to the left side such that:

`(1/2)(e_o)(x/2 - 1) = (3x)/2 - 3`

You need to factor out 3 to the left side such that:

`(1/2)(e_o)(x/2 - 1) = 3(x/2 - 1)`

You need to reduce duplicate factors such that:

`(1/2)(e_o) = 3 => e_o = 6`

Notice that the identity element for addition is `e_+ = 0` and the identity element for multiplication is `e_* = 1` .

Since the problem provides the information that `f(x) = ax + b` is an isomorphism of groups, hence, the following equations hold, such that:

`{(f(0) = 2),(f(1) = 6):} => {(a*0 + b = 2),(a*1 + b = 6):}`

`{(b = 2),(a +2 = 6):} => {(b = 2),(a = 6 - 2):} => {(b = 2),(a = 4):}`

**Hence, evaluating the coefficients a and b yields `a = 4` and `b = 2,` thus `f(x) = 4x + 2` .**