Mathematics, 18.03.2021 01:50 paralaw61772

20. Give an example of a function from N to N that is a) one-to-one but not onto. b) onto but not one-to-one. c) both onto and one-to-one (but different from the identity function). d) neither one-to-one nor onto.

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Answer from: NetherisIsTheQueen

Step-by-step explanation:

a) To provide an example of a function N → N that is one-to-one but not onto.

Suppose $f:N\to N$ to be f(n)=n^2

Then; $\text{a function } \ f: A \to B\ \text{is one-to-one if and only if } f(a) = f(b) \implies a = b \ for \ a, b \ \epsilon \ A.$

$\text{a function } \ f: A \to B\ \text{is onto if and only if for every element } b \ \epsilon \ B \\ \text{there exist an element a} \ \epsilon\ A \ such \ that f(a) = b}$

Now, assuming $a \ \Big {\varepsilon} \ N \& \ b \ \epsilon \ N;$

Then f(a) = f(b)

$a^2 = b^2 \\ \\ a = b$

The above function is said to be one-to-one

$\text{it is equally understandable that not every natural number is the square of a natural number}$ e.g

2 is not a perfect square, hence, it is not regarded as the image of any natural no.

As such, f is not onto.

We can thereby conclude that the function f(n) = n^2 is one-to-one but not onto

$Suppose f: N \to N$ be

$f(n) = [n/2] \\ \\ For \ n =1, f(1) = [1/2] = [0.5] = 1 \\ \\ For \ n=2 , f(2) = [2/2] = [1] = 1$

It implies that the function is not one-to-one since there exist different natural no. having the same image.

So, for $n \epsilon N$ , there exists an image of 2n in N

i.e.

f(2n) = [2n/2] = [n] = n

Hence, the function is onto

We thereby conclude that the function $f(n) = [n/2] \text{ is onto but not one-to-one}$

$let f: N\to N$ be $f(n) = \left \{ {{n+1, \ if \ n \ is \ even } \atop n-1 , \ if \ n \ is \ odd} \right.$

So, if n, m is odd:

Then:

$f(n) = f(m) \\ \\ n-1 = m-1 \\ \\ n = m$

Likewise, if n, m is even:

Then;

$f(n) = f(m) \\ \\ n+1 = m+ 1 \\ \\ n = m$

The function is then said to be one-to-one.

However, For $n \epsilon N$ and is odd, there exists an image of n - 1 that is even;

f(n - 1) = n -1 + 1 =n

For $n \epsilon N$ and is even, there exists an image of n + 1 that is odd;

f(n - 1) = n +1 - 1 = n

where(; implies such that)

Hence, this function is said to be onto.

We can therefore conclude that the function $f(n) = \left \{ {{n+1, \ if \ n \ is \ even } \atop n-1 , \ if \ n \ is \ odd} \right.$ is both onto and one-to-one.

Here, to provide an example where the $f:N \to N$ is neither one-to-one nor onto.

SO;

Let $f : N \to N$ is defined to be f(n)=0

Then, since every integer has the same image as zero(0), the function is not one-to-one.

Similarly, the function is not onto since every positive integer is not an image of any natural number.

We, therefore conclude that, the function f(n)=0 is neither one-to-one nor onto.

Answer from: Quest

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.

Answer from: Quest

i believe the answer is 323 ft

step-by-step explanation: