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3 years ago

10

Question 5 (5 points) Solve 5(8 - 2) + g = 6(8 - 4). a.0 b. all numbers c.0 d.2

1 answer:

anzhelika [568]3 years ago

8 0

By solving the equation i got g=-6 but i’m probs wrong

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Suppose that P(n) is a propositional function. Determine for which nonnegative integers n the statement P(n) must be true if a)

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Solution :

a). $P(0)$ is true

Then , $P(0+2)=P(2)$ is true.

$P(2+2)=P(4)$ is true

$P(4+2)=P(6)$ is true.

Therefore, we see that $P(n)$ is true for all the even integers : $\{0, 2,4,6,...\}$

b). $P(0)$ is true

Then , $P(0+3)=P(3)$ is true.

$P(3+3)=P(6)$ is true

$P(6+3)=P(9)$ is true.

Therefore, we see that $P(n)$ is true for all the multiples of 3 : $\{0, 3,6,9,12,...\}$

c). $P(0)$ and $P(1)$ is true, then $P(0+2)=P(2)$ is true

$P(1)$ and $P(2)$ is true, then $P(1+2)=P(3)$ is true.

$P(2)$ and $P(3)$ is true, then $P(2+2)=P(4)$ is true.

So, we observe that $P(n)$ is true for all the non- negative integers : $\{0, 1,2,3,4,5,6,...\}$ .

d). $P(0)$ is true,

So, $P(0+2)$ and $P(0+3)$ is true or $P(2)$ and $P(3)$ is true.

Now, $P(2)$ is true.

Again, $P(2+2)$ and $P(2+3)$ is true or $P(4)$ and $P(5)$ is true.

Now, $P(3)$ is true.

Again, $P(3+2)$ and $P(3+3)$ is true or $P(5)$ and $P(6)$ is true.

Thus,

$P(n)$ is true for all the non- negative integers except 1 : $\{0, 2,3,4,5,6,...\}$ .

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3 years ago