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## Saturday, 3 September 2011

### Math GRE - #32

Let $\mathbb{R}$ be the set of real numbers and let $f$ and $g$ be functions from $\mathbb{R}$ into $\mathbb{R}$. Which of the following is the negation of the statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

1. For each $s$ in $\mathbb{R}$, there does not exist an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$.
2. For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
3. There exists an $s$ in $\mathbb{R}$ such that for each $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
4. There exists an $s$ in $\mathbb{R}$ and there exists an $r$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.
5. For each $r$ in $\mathbb{R}$, there exists an $s$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.

Solution :

If you've got an innate talent for logic, you could probably tell that choice 3 is the answer after a bit of thinking.

Another way is to use quantifiers to translate the statement into formal logic.
The statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

can be translated as: $\forall s\in\mathbb{R},\;\exists r\in\mathbb{R},\; f(r)>0\rightarrow g(s)>0.$

We can negate this statement (imagine the negation as an arrow, flipping each quantifier as it flies past it) to be: $\exists s\in\mathbb{R},\;\forall r\in\mathbb{R},\; f(r)>0\wedge g(s)\leq0.$

This is choice 3 precisely.

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