Pages

News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!

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 :


Choice 3 is the answer.

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.

0 comments:

Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.