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balandron [24]
4 years ago
15

Write an integer whose absolute value is greater than itself.

Mathematics
1 answer:
Fed [463]4 years ago
8 0
<span>The set of an integer is called the set Z
The set of integers number is composed by the positive whole number {0, 1, 2, 3…} and the negative whole number {…-4, -3, -2, -1, 0} .We write it as follow Z= {…- 4, -3, -2, -1, 0}U{0, 1, 2, 3…}. If we assume that k is inside Z, so k can be a negative integer of positive integer. Besides the absolute value of a number is always the positive value of this number. Let be k<0 an integer different from zero, so abs(k) >0, and abs(k)> k for all value of k. For example for k= -2, and abs(k)=abs(-2)=2> -2 (always)</span>
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1. If a, b, c, d, and e are whole numbers and a(b(c
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a CANNOT be even ⇒ answer A

Step-by-step explanation:

* Lets revise the rules of even and odd numbers

- Even numbers any number its unit digit is (0 , 2 , 4 , 6 , 8)

- Odd numbers any number its unit digit is (1 , 3 , 5 , 7 , 9)

# even + even = even ⇒ 2 + 4 = 6  

# odd + odd = even ⇒ 1 + 3 = 4  

# odd + even = odd ⇒ 1 + 2 = 3  

# even × even = even ⇒ 2 × 4 = 8

# odd × odd = odd  

⇒ 3 × 5 = 15

# odd × even = even  ⇒ 5 × 6 = 30

∵ a[b(c + d) + e] = odd

∵ odd × odd = odd  

∴ a must be odd and [b(c + d) + e] must be odd

∵ odd + even = odd

∵ odd × even = even

# Case 1

∴ b(c + d) must be odd  if e even

∵ b(c + d) is odd

∴ b must be odd and (c + d) must be odd

∵ c + d must be odd

∵ odd + even = odd

∴ c or d can be even

- We now now that e , c and d can be even

# case 2

∴ b(c + d) must be even  if e odd

∵ b(c + d) is even

∵ even × even = even

∴ b and (c + d) both can be even

∵ c + d can be even

∴ c or d can be even or odd

- We now now that e , c , d and b can be even

∴ Only a can not be even

* a CANNOT be even

5 0
3 years ago
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