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Kipish [7]
3 years ago
8

What are the domain and range of the function represented by the set of ordered pairs? {(-7,1),(-3,2), (0, -2), (5,-5)}​

Mathematics
1 answer:
zimovet [89]3 years ago
7 0

Answer:

domain:  {-7, -3, 0, 5}

range:  {-5, -2, 1, 2}

Step-by-step explanation:

The domain of this set of ordered pairs is the set of input values (x-coordinates):  {-7, -3, 0, 5}, and the range is that of output values (y-coordinates):  {1, 2, -2, -5} or {-5, -2, 1, 2}.

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Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp
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Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If \phi: G \longrightarrow G is an homomorphism, then it must hold

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but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)

which tells us that \phi is a homorphism. The kernel of \phi

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n=3, we have

kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}

(c) If z_1,z_2 \in \mathbb{C}^{\times} remeber that

|z_1 \cdot z_2|=|z_1|\cdot|z_2|, which tells us that \phi is a

homomorphism. In this case

kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}, if we write a

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kern(\phi) \neq \{1\} the mapping is not 1-1, also if we take a negative

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\phi is not surjective.

(d) Remember that e^{ix}=\cos(x)+i\sin(x), using this, it holds that

\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)

which tells us that \phi is a homomorphism. By computing we see

that  kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \} and

Im(\phi) is the unit circle, hence \phi is neither injective nor

surjective.

7 0
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7 0
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