give answer in 2 step with explanation at the end of each step and final answer at the end:

Which of the following functions are one
-
to
-
one, onto, or both? Briefly explain your answer for f

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:

Which of the following functions are one
-
to
-
one, onto, or both? Briefly explain your answer for full marks. 
(
 
2
×
5
=
1
0
 ﻿points
)
Assume: 
x
,
yinZ, where 
Z
 ﻿is the set of all integer numbers.
a
)
 
y
a
=
b
s
 
(
x
)
where abs 
=
 ﻿absolute value of x
b
)
 
y
=
f
(
x
)
where 
f
(
x
)
=
 ﻿number of non
-
zero digits in x 
.
 ﻿For example: 
f
(
1
2
8
0
0
)
=
3
,
f
(
1
0
0
0
9
)
=
2
 ﻿etc.5 = ED)
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VIPSolutions · Accepted Answer

Let's answer both parts **step by step** as requested, with explanations and the final answer at the end.

---

### **a) $ y = \text{abs}(x) $, $ x \in \mathbb{Z} $**
#### **Step 1: One-to-one (Injective)?**
- **Explanation:**  
  A function is one-to-one if different inputs always give different outputs.
  - Here, $\text{abs}(2) = 2$ and $\text{abs}(-2) = 2$.
  - So, two different inputs ($2$ and $-2$) give the same output.
- **Conclusion:**  
  **Not one-to-one**.

#### **Step 2: Onto (Surjective)?**
- **Explanation:**  
  A function is onto if every possible output (in the codomain) is covered.
  - The codomain is $\mathbb{Z}$, all integers, but $\text{abs}(x)$ can never be negative.
  - So, for any negative integer $y$, there is no $x$ such that $\text{abs}(x) = y$.
- **Conclusion:**  
  **Not onto**.

#### **Final Answer for (a):**
> The function $ y = \text{abs}(x) $ is **neither one-to-one nor onto** (over $ \mathbb{Z} \to \mathbb{Z} $).

---

### **b) $ y = f(x) $, where $ f(x) = $ number of non-zero digits in $ x $**
#### **Step 1: One-to-one (Injective)?**
- **Explanation:**  
  A function is one-to-one if different inputs always give different outputs.
  - Example: $ f(12800) = 3 $ and $ f(12003) = 3 $.
  - Different numbers can have the same number of non-zero digits.
- **Conclusion:**  
  **Not one-to-one**.

#### **Step 2: Onto (Surjective)?**
- **Explanation:**  
  A function is onto if every possible output (in the codomain) is covered.
  - For any positive integer $ n $, there exists an $ x $ with exactly $ n $ non-zero digits (e.g., $ 111...1 $ ($ n $ times)).
  - But the output can **never** be negative or zero (except for $ x = 0 $, where $ f(0) = 0 $).
  - However, for all non-negative integers, you can find an $ x $ that matches.
- **Conclusion:**  
  The function is **onto** for codomain $\mathbb{Z}_{\ge 0}$ (non-negative integers), but **not onto** all $\mathbb{Z}$.

#### **Final Answer for (b):**
> The function $ y = f(x) $, with $ f(x) $ the number of non-zero digits in $ x $, is **not one-to-one** and **not onto** (if codomain is $\mathbb{Z}$), but **onto** if codomain is $\mathbb{Z}_{\ge 0}$.

---

## **Summary Table**

| Function         | One-to-One? | Onto (over $\mathbb{Z}$)? |
|------------------|-------------|-----------------------------|
| a) $ \text{abs}(x) $ | No          | No                          |
| b) $ f(x) $    | No          | No (onto $\mathbb{Z}_{\ge 0}$)  |

Let me know if you need explanations for a different codomain or more details!

VIPSolutions · Answer

Let's analyze both functions step by step, determining whether they are one-to-one (injective), onto (surjective), or both, with clear explanations.

---

### **a) $ y = |\mathbf{x}| $ where $ x \in \mathbb{Z} $**

#### **Step 1: Is it one-to-one (injective)?**
- **Explanation:**  
  A function is injective if different inputs always produce different outputs.
  - For example, $ |\mathbf{2}|=2 $ and $ |-\mathbf{2}|=2 $.  
  - Since $ x=2 $ and $ x=-2 $ produce the same output, the function is **not injective**.
- **Conclusion:**  
  **Not one-to-one**.

#### **Step 2: Is it onto (surjective)?**
- **Explanation:**  
  A function is surjective if every element in the codomain (here, $\mathbb{Z}$) has a pre-image.
  - The absolute value $ |\mathbf{x}| $ is always non-negative, so it can only produce outputs in $\{0, 1, 2, 3, ...\}$.
  - Negative integers in $\mathbb{Z}$ cannot be obtained as output.
- **Conclusion:**  
  **Not onto** (since negative integers are not covered).

---

### **Final conclusion for (a):**
> The function $ y=|\mathbf{x}| $ is **neither one-to-one nor onto** over $\mathbb{Z} \to \mathbb{Z}$.

---

### **b) $ y = f(\mathbf{x}) $, where $ f(x) $ = number of non-zero digits in $ x $**

#### **Step 1: Is it one-to-one (injective)?**
- **Explanation:**  
  Different inputs can have the same number of non-zero digits.
  - For example, $ f(12800) = 3 $ (digits '1', '2', '8') and $ f(12003) = 3 $ (digits '1', '2', '3').
  - These are different numbers but produce the same output.
- **Conclusion:**  
  **Not one-to-one**.

#### **Step 2: Is it onto (surjective)?**
- **Explanation:**  
  For the function to be onto, every integer in the codomain must be attainable.
  - Since $ f(x) $ counts non-zero digits, it can produce any non-negative integer (0, 1, 2, 3, ...).
  - For example:
    - $ f(0) = 0 $,
    - $ f(10) = 1 $,
    - $ f(101) = 2 $,
    - $ f(1002) = 3 $, etc.
  - Negative integers cannot be outputs because the count cannot be negative.
- **Conclusion:**  
  **Onto** if the codomain is the set of **non-negative integers** ($\mathbb{Z}_{\ge 0}$), but **not onto** if the codomain is all of $\mathbb{Z}$.

---

### **Final conclusion for (b):**
> The function $ y = f(x) $ (number of non-zero digits) is **not one-to-one** and **onto** the set $\mathbb{Z}_{\ge 0}$, but **not onto** the entire $\mathbb{Z}$.

---

### **Summary:**

| Function | One-to-One? | Onto? (over $\mathbb{Z}$) | Notes |
|----------|--------------|----------------------------|--------|
| a) $ |\mathbf{x}| $ | No | No | Absolute value is not injective and only covers non-negative integers |
| b) $ f(x) $ (non-zero digit count) | No | Yes (if codomain is $\mathbb{Z}_{\ge 0}$) | Counts non-zero digits, covers all non-negative integers |

**Note:** The analysis depends on the chosen codomain. Here, assuming the codomain is all integers ($\mathbb{Z}$), the functions are generally neither onto nor one-to-one, except for the non-zero digit function being onto $\mathbb{Z}_{\ge 0}$.

---

Let me know if you'd like further clarification!

VIPSolutions · Answer

Let's analyze both functions step by step, determining whether they are one-to-one (injective), onto (surjective), or both, with clear explanations.

---

### **a) $ y = |\mathbf{x}| $ where $ x \in \mathbb{Z} $**

**Step 1: Is it one-to-one (injective)?**  
- **Explanation:**  
  A function is injective if different inputs produce different outputs.  
  - For example, $ |\mathbf{2}|=2 $ and $ |-\mathbf{2}|=2 $.  
  - Since $ x=2 $ and $ x=-2 $ give the same output, the function is **not injective**.  
- **Conclusion:**  
  **Not one-to-one**.

**Step 2: Is it onto (surjective)?**  
- **Explanation:**  
  A function is surjective if every element in the codomain (here, $\mathbb{Z}$) has a pre-image.  
  - The absolute value $ |\mathbf{x}| $ is always non-negative, so it can only produce outputs in $\{0, 1, 2, 3, ...\}$.  
  - Negative integers in $\mathbb{Z}$ cannot be obtained as outputs.  
- **Conclusion:**  
  **Not onto**.

---

### **Final conclusion for (a):**  
> The function $ y=|\mathbf{x}| $ is **neither one-to-one nor onto** over $\mathbb{Z} \to \mathbb{Z}$.

---

### **b) $ y = f(\mathbf{x}) $, where $ f(x) $ = number of non-zero digits in $ x $**

**Step 1: Is it one-to-one (injective)?**  
- **Explanation:**  
  Different inputs can have the same number of non-zero digits.  
  - For example, $ f(12800) = 3 $ (digits '1', '2', '8')  
  - and $ f(12003) = 3 $ (digits '1', '2', '3')  
  - These are different numbers but produce the same output.  
- **Conclusion:**  
  **Not one-to-one**.

**Step 2: Is it onto (surjective)?**  
- **Explanation:**  
  For the function to be onto, every integer in the codomain must be attainable.  
  - Since $ f(x) $ counts non-zero digits, it can produce any non-negative integer (0, 1, 2, 3, ...).  
  - Examples:  
    - $ f(0) = 0 $  
    - $ f(10) = 1 $  
    - $ f(101) = 2 $  
    - $ f(1002) = 3 $  
  - Negative integers cannot be outputs because the count cannot be negative.  
- **Conclusion:**  
  **Onto** if the codomain is the set of **non-negative integers** ($\mathbb{Z}_{\ge 0}$), but **not onto** the entire $\mathbb{Z}$.

---

### **Final conclusion for (b):**  
> The function $ y = f(x) $ (number of non-zero digits) is **not one-to-one** and **onto** the set $\mathbb{Z}_{\ge 0}$, but **not onto** the entire $\mathbb{Z}$.

---

### **Summary:**

| Function | One-to-One? | Onto? (over $\mathbb{Z}$) | Notes |
|----------|--------------|----------------------------|--------|
| a) $ |\mathbf{x}| $ | No | No | Absolute value is not injective and only covers non-negative integers |
| b) $ f(x) $ (non-zero digit count) | No | Yes (if codomain is $\mathbb{Z}_{\ge 0}$) | Counts non-zero digits, covers all non-negative integers |

**Note:** The analysis assumes the codomain is all integers ($\mathbb{Z}$). If the codomain is the non-negative integers ($\mathbb{Z}_{\ge 0}$), then the second function is onto but not one-to-one.

Let me know if you'd like further clarification!

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