Prove that for all real numbers $x$ and $y$, if $x+y \geq 100$, then $x \geq 50$ or $y \geq 50$.

Question

I'm confused about the following question in my math textbook.

Prove that for all real numbers $x$ and $y$, if $x+y \geq 100$, then $x \geq 50$ or $y \geq 50.$

The or is what gets me. For or to be true don't we need only one of the statements in the operation to be true? Couldn't we have $x = 12, y = 55, x+y = 67$ is there something I'm missing here? Shouldn't it be an and instead of an or?

In your example, since $x+y=67<100$, the premise “$x+y\geq 100$” is false. If the premise of an implicational statement is false, then the implicational statement is true. Formally, if $P$ is false, then the statement “$P$ implies $Q$” is true for any condition $Q$. Thus, your example is consistent with the claim; it is not a counterexample. Actually, what your example does show is that the converse statement “if $x\geq 50$ or $y\geq 50$, then $x+y\geq100$” is false! — triple_sec, Jul 05 '14 at 01:07
Note this if $x+y\geq 100$ then $x\geq 50$ or $y\geq 50$ can also be proved by proving "if $x<50$ and y<50, then $x+y<100$" — Kamster, Jul 05 '14 at 01:10
You can also just prove original statement by breaking it up into cases 1.) $x\geq 50$ and 2.) $x<50$ — Kamster, Jul 05 '14 at 01:14

Mark Fantini · Accepted Answer · 2014-07-05T01:10:45.037

8

The problem is with the connective 'or'. It is not (always) exclusive. Suppose you took $x=12$, then to satisfy the inequality $x+y\geq 100$ you need $y \geq 88$, which does satisfy $y \geq 50$.

Hint: Prove the contrapositive.

Edit: Answering the comment below, no, because $x+y=72 < 100$. Your hypothesis is that $x+y \geq 100$. In other words, you have to think in this order:

Pick numbers such that $x+y \geq 100$.
Check if $x \geq 50$.
Check if $y \geq 50$.

If 2 or 3 is verified then you strengthen your belief that it is true. Otherwise you have found a counterexample, but it is imperative that $x+y \geq 100$ holds.

edited Jul 05 '14 at 01:10

answered Jul 05 '14 at 01:04

Mark Fantini

5,523

Yes but doesn't $x = 12$ and $y= 60$ satisfy the equation because one of the operands is true, which is all an or operation needs to output a true value? – Dunka Jul 05 '14 at 01:07
Nevermind, brain fart. Thanks very much for your help. – Dunka Jul 05 '14 at 01:09
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I've edited it answering your comment. If it led you to the correct conclusions please consider accepting. =) Best wishes. – Mark Fantini Jul 05 '14 at 01:11

score 1 · Answer 2 · answered Jul 05 '14 at 01:31

The form of the statement that you need to prove is "if A then B". When you speak about the "or" in your statement, you are discussing whether B is true. But recall that "if A then B" does NOT imply that "if B then A". Let's apply that to your example...

Let $x,y\in\mathbb{R},\:x=12,\:y=55$.
Then $x+y=67\lt100$.
Now let A = "$x+y\geq100$" and B = "$x\geq50$ or $y\geq50$".
Given our values for $x$ and $y$, we have A is false and B is true.
So we have "if A then B" is true and "if B then A" is false.

It's still up to you to prove the statement, but we've shown why your example is not really a counter-example.

Michael Hardy · Answer 3 · 2014-07-05T02:04:53.460

1

"Before taking this course, you must have passed either a course on omphalology or a course on exorcism of monsters."

The "or" above clearly means one or the other or both.

It's standard in mathematical reasoning to construe "or" that way.

edited Jul 05 '14 at 02:04

answered Jul 05 '14 at 01:52

Michael Hardy

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I'm very interested in taking any of those. Do you know of any MOOCs? =) – Mark Fantini Jul 05 '14 at 01:59
@Fantini : Certainly. – Michael Hardy Jul 05 '14 at 02:02
I take it no one here has heard of the Dan Savage doctrine. A much healthier attitude than assuming that anyone who asks a basic question must be a troll. – Robert Soupe Jul 05 '14 at 02:02
I don't think Michael considered the asker a troll but instead wanted to make a joke. At least, that's how I interpreted it. – Mark Fantini Jul 05 '14 at 02:04
Does anyone think those asking a basic question is a troll? My post answers the question about the meaning of "or". – Michael Hardy Jul 05 '14 at 02:04
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OK, I've changed it to "monsters". – Michael Hardy Jul 05 '14 at 02:05
College of omphalology: http://www.bmj.com/content/3/5872/176.3 – Michael Hardy Jul 05 '14 at 02:06
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"omphalos" is Greek for belly button, and probably omphalology would be the medical specialty that treats thereof. – Michael Hardy Jul 05 '14 at 02:06
@MichaelHardy Yes, lots of people on Yahoo! Answers think that way, and probably a few on this website, too. – Robert Soupe Jul 05 '14 at 04:37
@RobertSoupe : I have no idea what you mean by "the Dan Savage doctrine" nor what your second comment here means. – Michael Hardy Jul 05 '14 at 16:10
@MichaelHardy Dan Savage is a sex advice columnist and people ask him all the time if the letters he gets are real, for, after all, it's possible for someone to write down something that never happened and send it to him. But he never worries about publishing a letter from a troll. – Robert Soupe Jul 05 '14 at 17:24
@RobertSoupe : OK. But I never suspected a troll was involved here. – Michael Hardy Jul 06 '14 at 15:03
@MichaelHardy Ah, so I got ensnared in the OR trap same as the OP. – Robert Soupe Jul 06 '14 at 16:58

score 1 · Answer 4 · answered Jul 05 '14 at 01:59

There is no mistake, but maybe they could have worded it better, or maybe you haven't yet learned that converses don't always hold.

For example, "If an animal is a dog, then that animal is a canine." That's true, right? But does the converse hold? "If an animal is a canine, then that animal is a dog." That's not necessarily true. The animal could be a wolf, a coyote, a fox, etc.

So, "If $x + y \geq 100$, then either $x \geq 50$ or $y \geq 50$." We could have $x = 3^7$ and $y = (-7)^3$, and $x + y$ is still more than 100. Suppose $x = y = 7^2$. Then $x + y$ falls short of 100, because both $x$ and $y$ fall short of 50.

The converse would be "If $x \geq 50$ or $y \geq 50$, then $x + y \geq 100$." But as you have already demonstrated, this is not always true. The condition "$x \geq 50$ or $y \geq 50$" is not equivalent to the condition "$x + y \geq 100$."

The converse of a true statement is also true if and only if the statement and its converse are equivalent. For example, "If a positive integer is the square of a prime, then it has exactly three positive divisors." Its converse is "If a positive integer has exactly three positive divisors, then it is the square of a prime." Both statements are true and they are equivalent.

score 1 · Answer 5 · answered Jul 05 '14 at 01:59

English language, unfortunately, does not explicitly distinguish between the inclusive (both can be true) and exclusive (both can't be true) meanings of the word "or".

In formal logic, $\vee$ (often pronounced "or") is always inclusive, and consequently, in mathematics, the word "or" is usually meant inclusively, and that is what is meant in this case.

Although one should keep in mind that when we speak in English, we might wind up using "or" exclusively, even if we are talking mathematics. Unfortunately, one must be prepared to infer from context which variant of "or" is meant. :(

Prove that for all real numbers $x$ and $y$, if $x+y \geq 100$, then $x \geq 50$ or $y \geq 50$.

5 Answers5