Given f: N → N, given by f(x) = x2
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
x2 = y2
x = y (We do not get ± because x and y are in N that is natural numbers)
So, f is an injection.
Surjection condition:
Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).
f(x) = y
x2= y
x = \( \sqrt{y}\), which may not be in N.
For example, if y = 3,
x = \( \sqrt{3}\) is not in N.
So, f is not a surjection.
Also f is not a bijection.
Answered by Aaryan | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y
Express R and R-1 as sets of ordered pairs. Determine also
(i) the domain of R‑1
(ii) The Range of R.
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