What is the proper way to write this statement mathematically? "All people are smokers or non-smokers but not both."

$H=humans,S=smoker,NS=non-smoker$

Raegan Bray
2022-07-15
Answered

Properly Translate Propositional Statement to Discrete Math

What is the proper way to write this statement mathematically? "All people are smokers or non-smokers but not both."

$H=humans,S=smoker,NS=non-smoker$

What is the proper way to write this statement mathematically? "All people are smokers or non-smokers but not both."

$H=humans,S=smoker,NS=non-smoker$

You can still ask an expert for help

Seromaniaru

Answered 2022-07-16
Author has **12** answers

Step 1

How do you represent the "not both" part? You don't. In the statement you wrote, x could be both S and NS and it would be true. You need this addition:

$\mathrm{\forall}x\in H:(x\in S\vee x\in NS)\wedge \mathrm{\neg}(x\in S\wedge x\in NS)$

Step 2

The confusing thing with this problem is that "smoker" and "non-smoker" are mutually exclusive sets in natural language. They are also mutually exclusive in math if you define NS as $\mathrm{\neg}S$, which would be an intuitive and absolutely correct thing to do in a different context. But in the context of this problem, we do not have this definition explicitly. In fact, because the problem explicitly says "but not both" we are guided to choose S and NS as two distinct sets, as you have done.

How do you represent the "not both" part? You don't. In the statement you wrote, x could be both S and NS and it would be true. You need this addition:

$\mathrm{\forall}x\in H:(x\in S\vee x\in NS)\wedge \mathrm{\neg}(x\in S\wedge x\in NS)$

Step 2

The confusing thing with this problem is that "smoker" and "non-smoker" are mutually exclusive sets in natural language. They are also mutually exclusive in math if you define NS as $\mathrm{\neg}S$, which would be an intuitive and absolutely correct thing to do in a different context. But in the context of this problem, we do not have this definition explicitly. In fact, because the problem explicitly says "but not both" we are guided to choose S and NS as two distinct sets, as you have done.

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Prove that if $A\subseteq B$, and A is uncountable, then B is uncountable"

I think that for an answer we could reason with cardinality as follows Suppose that B is countable, then $\mid B\mid \le \mid \mathbb{N}\mid $.

But if $A\subseteq B$, then $|A|\le |B|\le |\mathbb{N}\mid $ that is untrue if A is uncountable, since any uncountable set should have higher cardinality then natural numbers.

I think that for an answer we could reason with cardinality as follows Suppose that B is countable, then $\mid B\mid \le \mid \mathbb{N}\mid $.

But if $A\subseteq B$, then $|A|\le |B|\le |\mathbb{N}\mid $ that is untrue if A is uncountable, since any uncountable set should have higher cardinality then natural numbers.