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Let R be set of all real numbers, and

A = B = R*R

A function A-> B is defined by

f(a,b) = (a+b,a-b)

How to prove it is a bijective function?
in Set Theory & Algebra by Junior (853 points) | 67 views
You have to prove that sum and difference of 2 real numbers are always unique.
To prove a function is bijective, prove that it is one-one and onto i.e for every value of (a,b) there exists a unique value of f(a,b) and that for every value (x,y) present in the codomain, there exists a value (a,b) such that f(a,b) = (x,y).

1 Answer

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Let's there is a value $(x, y)$  in co-domain of $RXR$. From there we can get $a=(x+y)/2$ & $b=(x-y) /2$ and $a, b \epsilon R$. Thus proves Range = Co domain... This also proves for every $(x, y)$, there is unique pair or $(a, b) $. Thus proves function is bijective.
by Active (1.9k points)

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