Logic: If $\frac{1}{x} > 0.5$, then $x$ can be any real number smaller than $2$?

Question

My friends are arguing whether the following statement is correct or not. It seems easy enough but I cannot tell who's argument is correct.

If $\frac{1}{x} > 0.5$, then $x$ can be any real number smaller than $2$.

We all agree that the range of $x$ should be found as follows. $$\frac{1}{x} > 0.5$$ $$x^2 \times \frac{1}{x} > x^2 \times 0.5$$ $$0>0.5x^2 - x$$ $$0<x<2$$

However, they have a different conclusion afterwards.

Argument $1$:

The range is suggesting that $x$ cannot be negative. Hence, the word 'any real numer' is wrong . The statement is therefore false.

Argument $2$:

In order for the 'if' statement to be true, we let $f(x) = \frac1x$ and we can already ignore the case in which $x$ is negative and assume that the domain is the set of all positive real numbers. In this domain, all numbers must be smaller than $2$ for $f(x)$ to be smaller than $2$. Hence, the statement is true.

Can you share your view on this? Is the statement true or false? Thank you.

Answer 1

The statement to (dis)prove is of the form: "If A , Then B."

Since in your case, statement A is equivalent to the result $0 < x < 2$, we can rewrite it as:

"If $\:0<x<2$, Then $x$ can be any real number smaller than 2", which is clearly false, as $x$ cannot be, for example, $-1$.

In a case like this, for the original statement to be true, we want the solutions to statement B to be a subset or equal to the solutions to statement A, rather than a superset.

For example, a statement that would work would be: "If $\frac{1}{x} < 0.5$, Then $0<x<2$"

Answer 2

My friends are arguing whether the following statement is correct or not.

If $\frac{1}{\boldsymbol x} > 0.5,$ then $\boldsymbol x$ can be any real number smaller than $2.$

Unless $x$ symbolises a constant—in which case the statement is false because $x$ is not any but some (in fact, a particular) real number smaller than $2$—the quoted line is not even a statement, due to $x$ being a free variable.

On the other hand, this is a false statement:

• for each $\boldsymbol x,$ if $\frac{1}{\boldsymbol x} > 0.5$, then $\boldsymbol x$ can be any real number smaller than $2.$

And these are true statements:

• for some $\boldsymbol x,$ if $\frac{1}{\boldsymbol x} > 0.5$, then $\boldsymbol x$ can be any real number smaller than $2$
• for each $\boldsymbol x,$ if $\frac{1}{\boldsymbol x} > 0.5$, then $\boldsymbol x$ is a real number smaller than $2.$

Argument $1$:

The range is suggesting that $x$ cannot be negative. Hence, the word 'any real number' is wrong . The statement is therefore false.

No:

• if $\frac{1}{\boldsymbol x} > 0.5,$ then $\boldsymbol x$ can be any real number strictly between $0$ and $2$

is still not a statement, while

• for each $\boldsymbol x,$ if $\frac{1}{\boldsymbol x} > 0.5$, then $\boldsymbol x$ can be any real number strictly between $0$ and $2$

is still a false statement.

Argument $2$:

In order for the 'if' statement to be true, we let $f(x) = \frac1x$ and we can already ignore the case in which $x$ is negative and assume that the domain is the set of all positive real numbers. In this domain, all numbers must be smaller than $2$ for $f(x)$ to be smaller than $2$. Hence, the statement is true.

No, if we restrict the domain of discourse to positive numbers and treat the quoted line as an implicitly universally quantified statement, then the statement is false, because each instantiation of $x$ (i.e., ‘for each $x\text’)$ is still a single real number rather than any real number.

On a separate note: given a number $c$ such that $0<c<2,$ surely you agree that it is perfectly correct to assert that $c<2$ ? After all, we are claiming merely that the former inequality implies the latter inequality, not that they are equivalent to each other.