2

I have this statement:

It can be assured that | p | ≤ 2.4, if it is known that:

(1) -2.7 ≤ p <2.3

(2) -2.2 < p ≤ 2.6

My development was:

First, $ -2.4 \leq p \leq 2.4$

With $1)$ by itself, that can't be insured, same argument for $2)$

Now, i will use $1)$ and $2)$ together, and the intersection between this intervals are: $(-2.2, 2.3)$. So, this also does not allow me to ensure that | p | ≤ 2.4, since, there are some numbers that are outside the intersection of these two intervals, for example 2.35 is outside this interval.

But according to the guide, the correct answer must be $1), 2)$ together. And i don't know why.

Thanks in advance.

ESCM
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3 Answers3

2

You getting confused by a strong premise implying a weak conclusion.

1 and 2 together say precisely: $p\in (-2.2,2.3) $

And you are asking to conclude $p\in [-2.4,2.4] $.

That's simply a matter of noting if $(-2.2, 2.3) \subset [-2.4,2.4] $. And if $p $ is in a smaller subset, then we can conclude $p $ must therefore also be in the bigger superset.

Or in other words:

$-2.2 < p < 2.3\implies $

$-2.4\le -2.2 < p <2.3 \le 2.4\implies $

$-2.4\le p \le 2.4$

....

An analogy:

We can conclude $p $ is a somewhat green article of clothing if we know both

1: $p$ is a hat.

2: $p$ is precisely the color an emerald takes when viewed at noon on the summer solstice on the equator.

In your question, 1 tells you that $p <2.3$ so you can conclude $p\le 2.4$. ($p $ is a hat so you can conclude $p $ is an article of clothing.)

And 2 tells you that $-2.2 <p $ so you can conclude $-2.4\le p $. ($p $ is the color of emeralds so you can conclude $p $ is green.)

Together 1 and 2 tell you $-2.2<p <2.3$ so you can conclude $-2.4\le p\le 2.4$. (Together you know $p $ is a hat the color of emeralds in a certain condition so you can conclude $p $ is a green article of clothing.)

Strong premise. Weak result.

fleablood
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  • I had thought about it. For example, if you AFFIRM that k <10, then I can affirm that: k <15. However, my problem is that, I think, I CAN NOT affirm that it is "lesser or EQUAL" than 15, since it can take the value 15 and that contradicts the initial statement of k <10. – ESCM Jun 01 '19 at 20:44
  • However, I think I understand the point. A number k can be between: 10 < k <100, and at the same time, I with that i can affirm that: -10000 <k <100000, regardless of whether there are intermediate values, this is still true, this is what I understood. – ESCM Jun 01 '19 at 20:52
  • The key word is "less than OR equal" is the "OR". If $5 < 16$ then $5 \le 16$. $5 \le 16$ does not mean that $5 =16$. It doesn't even mean $5$ might be less than $16$. It means that exactly one of the two statements: A) $5 < 16$ and B) $5=16$ is true. And one of them IS true. A) is true. So A OR B is true. – fleablood Jun 03 '19 at 14:47
  • We are asked to verify "Fred is green or red". We sneak up on Fred and verify that he is a solid green. SO we say "Yep, Fred is green so the statement is true: Fred is green or red". But then one of us says. "But he isn't red. So he can't be green OR RED because he isn't red". Well... the first person is right. Fred is completely green. That is a subset of being green OR red. So GREEN $\implies$ GREEN or RED. – fleablood Jun 03 '19 at 14:51
  • I understand the point. Where i can get info about " strong premise implying a weak conclusion." – ESCM Jun 04 '19 at 19:23
0

$ -2.4 \leq p \leq 2.4$

$ -2.7 \leq p < 2.3$

$-2.2 < p \leq 2.6 $

Your answer is common intersection of these inequalities.

$ p \in (-2.2, 2.3)$

Vineet
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0

The point is that if you use $1$ and $2$ together you know $-2.2 \lt p \lt 2.3$. This does allow you to ensure $|p| \le 2.4$ because all numbers in the interval are less than $2.4$ in absolute value. The fact that there are numbers outside the interval that also qualify does not matter at all. If I told you that $0.1 \le p \le 0.9$ you could assure me that $|p| \le 2.4$. You could also make a stronger statement, like $|p| \le 1$ or $|p| \le 0.9$, but that doesn't mean the one with $2.4$ is incorrect.

Ross Millikan
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