Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.
Answers
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Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z:
justify your answer.
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➡️Given relation f is defined as
➡️f = {(ab, a + b): a, b ∈ Z}
➡️We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.
➡️As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f
➡️i.e., (12, 8), (12, –8) ∈ f
➡️It’s clearly seen that, the same first element, 12 corresponds to two different images (8 and –8).
➡️Therefore, the relation f is not a function.
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Answer:
Here,
f = {ab,(a+b)}:a,b∈Z}
Let we take a = 0, b=1 then, ab =0 and (a +b) = 0 + 1 = 1 then, (0,1)∈f
Again, we take a =0, b =2
So, ab = 0*2= 0
a + b = 0 + 2 = 2
Then, (0,2)∈f
But a/c to condition of function it's not possible .(see the daigram).[ 1st set contains more then one image in another set ]
Hence, it's not a function.