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AveGali [126]
3 years ago
10

Give an example of a function fromNtoNthat isa)one-to-one but not onto.b)onto but not one-to-one.c)both onto and one-to-one (but

different from the iden-tity function).d)neither one-to-one nor onto.
Mathematics
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

Step-by-step explanation:

Let f be a function from N to N.

N_set of all natural numbers

i) one to one but not onto

consider the function

f(x) = x^2

When two numbers have same square we find that the numbers should be the same because they are positive.

So one to one but not onto because consider 3 it does not have square root in N.

ii) Onto but not one to one

Consider

f(x) = x, x odd\\f(x) = x/2, x even.

this is onto because every number has a preimage in N.

But not onto because consider 6 and 3, f(6) = 3 and f(3) =3

So not one to one

iii) both onto and one-to-one

f(x) = \\\\x-1,x odd

=x+1, x even

This is both one to one and onto since we consider only integers  

iv) Neither one to one nor onto

Consider the function

f(x) = 2

This is not onto because 3 cannot have a preimage in N, not one to one because f(1) = f(2) where 1 not equals 2

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Solving a system of equations, we will see that there are 6000 kg of coal on the pile.

<h3>How many kilograms of coal are in the pile of coal?</h3>

Let's say that there are N kilograms of coal, and it is planned to burn it totally in D days, these are our variables.

We know that:

  • If the factory burns 1500 kilograms per day, it will finish burning the coal one day ahead of planned.
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Then we can write the system of equations:

(1500)*(D - 1) = N

(1000)*(D + 1) = N

Because N is already isolated in both sides, we can write this as:

(1500)*(D - 1) = N = (1000)*(D + 1)

Then we can solve for D:

(1500)*(D - 1) =  (1000)*(D + 1)

1500*D - 1500 = 1000*D + 1000

500*D = 2500

D = 2500/500 = 5

Now that we know the value of D, we can find N by replacing it in one of the two equations:

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If you want to learn more about systems of equations:

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