GATE Overflow Test Series | Discrete Mathematics | Test 1 | Question: 24

Question

Suppose that $h:\mathbb{Z} \rightarrow \mathbb{Z}$ is known to be one-one and onto (bijection). Let's define a function $f:\mathbb{Z}^2 \rightarrow \mathbb{Z}$ by $f(x,y)=h(x)+h(y).$ Which of the following defines the nature of function $f(x,y)?$

• A. $f$ is one-one
• B. $f$ is onto
• C. $f$ is a bijection
• D. $f$ is neither one-one nor onto

Comments

gatecse sir, can u please tell that since in the question its mentioned that f(x,y) maps from Z^2  to Z . this means that only Z+ are going to get covered in co-domain not Z-.    this implies that the function is clearly NOT ONTO.  Also since the mapping is Z^2 to Z there is higher probability that function cannot be one to one.   Please correct me if I’m wrong 

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$Z^2$ means ordered pairs or tuples of the form (Z, Z) so $Z^-$ will get convered

Answer 1

For onto, we have to show that for every element in the co-domain set we have a value in the domain set (range = co-domain). Here, $h$ is given as onto and is from $\mathbb{Z}$ to $\mathbb{Z}.$ $f$ is from $\mathbb{Z}^2$ to $\mathbb{Z}$ and is defined as  $f(x,y)=h(x)+h(y).$ Since $h$ is onto, we have an inverse image for any value $y$ in $h(y).$ So, by taking $h(x)= 0 \implies x = h^{-1}(0),$ we get $f(x,y) = 0 + h(y).$ This will cover the entire $\mathbb{Z}$ and so, $f$ is onto.

$f$ is not one-one as $f(2,4)$ and $f(4,2)$ have the same value.

Comments

Suppose,

$h(x)=x$ be a one-one and onto function from $Z\rightarrow Z$.

Then for what values of x and y $f(x,y)=h(x)+h(y)$, negative numbers will be covered in the range of the function $f(x,y)$.

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@HeartBleed

Consider scenario where $x < y$ . Note that $x$ can take negative values as well since domain $Z$ is a set of integers

For e.g., $x=-6$, $y=2$. As per your definition of $h(x) = x$, we get

$h(x) = -6$

$h(y) = 2$

$h(x,y) = h(x) + h(y) = -6 + 2 = -4$ 

Hope this clarifies