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AleksandrR [38]
3 years ago
14

Determine whether the relation o defined on ℤ is reflexive, symmetric, or transitive. The relation o on ℤ is defined as follows:

for all m, n ∈ ℤ, m o n ⇔ m – n is odd
Mathematics
1 answer:
Alexxx [7]3 years ago
5 0
  • Reflexive: no. A relation is reflexive is an element is in relation with itself. But in this case, it cannot happen, because x \circ x \iff x-x \text{ is odd} \iff 0 \text{is odd}, which clearly cannot happen
  • Symmetric: yes. A relation is symmetric if every time x and y are in relation, then also y and x are in relation. In this case, you have x and y are in relation if x-y is odd. But then, this guarantees that y and x are in relation, because y-x = -(x-y), and the opposite of an odd number is still odd
  • Transitive: no. A relation is transitive if every time x and y are in relation, and y and z are in relation, then x and z are in relation. So, suppose that x and y are in relation, which means x-y = 2k+1, for some integer k. We also know that y and z are in relation, which means that x-y = 2m+1, for some integer m. But then, you have

x-z = x-y+y-z = 2k+1 + 2m+1 = 2k+2m+2 = 2(k+m+1)

But since 2(k+m+1) is twice some integer, it is even, and thus x and z are not in relation. So, we've proven that although x \circ y and y \circ z, it can't be that x \circ z, and thus the relation is not transitive.

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Plutonium–238 has a yearly decay constant of 7.9 × 10-3. If an original sample has a mass of 15 grams, how long will it take to
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The answer is 28 years

At = A0 * e^(-k * t)
At = 12 g
A0 = 15 g
k = 7.9 × 10^-3 = 0.0079 
t = ?

12 = 15 * e^(-0.0079 * t)
12/15 = e^(-0.0079 * t)
0.8 = e^(-0.0079 * t)

Logarithm both sides (because ln(e) = 1:
ln(0.8) = ln(e^(-0.0079 * t))
ln(0.8) = (-0.0079 * t) * ln(e)
-0.223 = -0.0079 * t
t = -0.223 / -0.0079
t = 28.23
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