in Set Theory & Algebra edited by
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1 vote
1 vote

I think only d) is correct

in Set Theory & Algebra edited by
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4 Comments

Real numbers can't be in one-to-one correspondence since it is uncountable.

For natural numbers and integers the co-domain is not equal to range.

Rational numbers will hold true for the function here..
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@Prince Sindhiya

again posting screenshot unnecessarily.

If you do this, we can't find duplicates, hence more redundancy in GO platform and Moreover our time will be wasted unnecessarily. I know to type it takes 10-15 min, but atleast for next time if someone ask the same question i can easily find this search

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@Prince Sindhiya 4 would be correct, 1 and 2 cannot but why not 3 ?

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edited by

f(12(x−1))=2(12(x−1))+1=(x−1)+1=x, therefore surjective.

Now suppose f(x1)=f(x2)

then f(x1)2x1+12x1x1=f(x2)=2x2+1=2x2=x2, therefore injective.

Ref: https://math.stackexchange.com/a/1696547/864381

I think this reasoning is applied for Rational numbers too. Therefor answer is (C).

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