Explain why multiplication by 2 defines a bijection from R to R but not from Z t

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a bijection must be injective AND surjective. often, the importance of being surjective is overlooked, because the co-domain and the image are assumed to be the same. here is how it plays out for R and Z: x-->2x is clearly injective for both: 2x = 2y implies x = y in both R and Z (although it is harder to prove this in Z, one has to use divisiblity arguments). but x-->2y is surjective in R, because we have the (two-sided) inverse: y-->y/2 (or, if you prefer, every real number y has the real pre-image y/2). in Z, the situation is not so good, there is no odd integer in the image of the map: x-->2x. or It's not a bijection in Z because it's not surjective. For example, f(x) = 3 has no solution in Z. In other words, you can't double an integer (Z) to get an odd number. It works in R because it's ok to have decimals

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