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Lunna [17]
3 years ago
14

If S is countable and nonempty, prove their exist a surjection g: N --> S

Mathematics
1 answer:
aleksley [76]3 years ago
8 0

Answer with Step-by-step explanation:

We are given S be any set which is countable and nonempty.

We have to prove that their exist a surjection g:N\rightarrow S

Surjection: It is also called onto function .When cardinality of domain set is greater than or equal to cardinality of range set then the function is onto

Cardinality of natural numbers set =\chi_0( Aleph naught)

There are two cases

1.S is finite nonempty set

2.S is countably infinite set

1.When S is finite set and nonempty set

Then cardinality of set S is any constant number which is less than the cardinality of set of natura number

Therefore, their exist a surjection from N to S.

2.When S is countably infinite set and cardinality with aleph naught

Then cardinality of set S is equal to cardinality of set of natural .Therefore, their exist a surjection from N to S.

Hence, proved

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3 years ago
Which equation is y = –3x2 – 12x – 2 rewritten in vertex form?
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Answer:

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Step-by-step explanation:

* Lets revise how to put the quadratic in the vertex form

- The general form of the quadratic is y = ax² + bx + c, where

 a , b , c are constants

# a is the coefficient of x²

# b is the coefficient of x

# c is the numerical term or the y-intercept

- The vertex form of the quadratic is a(x - h)² + k, where a, h , k

 are constants

# a is the coefficient of x²

# h is the x-coordinate of the vertex point of the quadratic

# k is the y-coordinate of the vertex point of the quadratic

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- We find k by substitute the value of h instead of x in the general form

 of the quadratic

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∵ y = -3x² - 12x - 2

∵ y = ax² + bx + c

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∵ h = -b/2a

∴ h = -(-12)/2(-3) = 12/-6 = -2

- Lets find k

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* Lets writ the vertex form

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∴ y = -3(x - -2)² + 10

∴ y = -3(x + 2)² + 10

* The vertex form is y = -3(x + 2)² + 10

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