Show `a/b+c/d=(ad+bc)/(bd)` :

`a/b+c/d` is given

Since `d/d=1` and `b/b=1` by the multiplicative inverse axiom we have:

`a/b+c/d=a/b * d/d + b/b*c/d` by the multiplicative identity axiom

`=(ad)/(bd)+(bc)/(bd)`

`=1/(bd)(ad+bc)` by the distributive axiom of mult. over addition

`=(ad+bc)/(bd)` as required.

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Show `a/b+c/d=(ad+bc)/(bd)` :

`a/b+c/d` is given

Since `d/d=1` and `b/b=1` by the multiplicative inverse axiom we have:

`a/b+c/d=a/b * d/d + b/b*c/d` by the multiplicative identity axiom

`=(ad)/(bd)+(bc)/(bd)`

`=1/(bd)(ad+bc)` by the distributive axiom of mult. over addition

`=(ad+bc)/(bd)` as required.