Chant6j

Answered

2022-07-05

How to solve the system of inequalities when using topological proof to show that there are exactly 5 platonic solids?

I have seen other answers explaining the topological proof up until the point of

$1/p+1/q>1/2$ and $p$, $q$ are greater than or equal to three

Then they proceed to say that the 5 platonic solids have the only values that satisfy these conditions. My question is how do I solve for these values?

I have seen other answers explaining the topological proof up until the point of

$1/p+1/q>1/2$ and $p$, $q$ are greater than or equal to three

Then they proceed to say that the 5 platonic solids have the only values that satisfy these conditions. My question is how do I solve for these values?

Answer & Explanation

Kiana Cantu

Expert

2022-07-06Added 22 answers

We're looking for integer values of $p$ and $q$ that satisfy the three given equations $p\ge 3$, $q\ge 3$ and $\frac{1}{p}+\frac{1}{q}>\frac{1}{2}$.

Suppose both $p$ and $q$ are not equal to $3$. Since they are integers, they must both be greater or equal to $4$. But if that's the case, then $\frac{1}{p}$ and $\frac{1}{q}$ are both less or equal to $\frac{1}{4}$, meaning that $\frac{1}{p}+\frac{1}{q}\le \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$, contradicting one of the equations.

So to solve the equation either $p$ or $q$ must be equal to $3$.

Let's consider only the case where $p$ is equal to $3$. Then we have $\frac{1}{3}+\frac{1}{q}>\frac{1}{2}$. Subtracting $\frac{1}{3}$ from both sides yields $\frac{1}{q}>\frac{1}{6}$, which implies $q<6$.

So if $p$ is equal to $3$, $q$ must be an integer with $3\le q<6$, so $q$ must be either $3$ or $4$ or $5$.

By a symmetrical argument, if $q$ is equal to $3$, $p$ must be equal to $3$ or $4$ or $5$.

So we have five cases remaining:

$(p=3,q=3),(p=3,q=4),(p=3,q=5),(p=4,q=3),(p=5,q=3).$

It is now easy (and necessary) to check that all these five cases satisfy the three given equations.

There isn't really a general method to do this kind of proof, it just requires finding a bunch of solutions, then a sufficient bunch of good arguments for why there aren't any more solutions.

Suppose both $p$ and $q$ are not equal to $3$. Since they are integers, they must both be greater or equal to $4$. But if that's the case, then $\frac{1}{p}$ and $\frac{1}{q}$ are both less or equal to $\frac{1}{4}$, meaning that $\frac{1}{p}+\frac{1}{q}\le \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$, contradicting one of the equations.

So to solve the equation either $p$ or $q$ must be equal to $3$.

Let's consider only the case where $p$ is equal to $3$. Then we have $\frac{1}{3}+\frac{1}{q}>\frac{1}{2}$. Subtracting $\frac{1}{3}$ from both sides yields $\frac{1}{q}>\frac{1}{6}$, which implies $q<6$.

So if $p$ is equal to $3$, $q$ must be an integer with $3\le q<6$, so $q$ must be either $3$ or $4$ or $5$.

By a symmetrical argument, if $q$ is equal to $3$, $p$ must be equal to $3$ or $4$ or $5$.

So we have five cases remaining:

$(p=3,q=3),(p=3,q=4),(p=3,q=5),(p=4,q=3),(p=5,q=3).$

It is now easy (and necessary) to check that all these five cases satisfy the three given equations.

There isn't really a general method to do this kind of proof, it just requires finding a bunch of solutions, then a sufficient bunch of good arguments for why there aren't any more solutions.

prirodnogbk

Expert

2022-07-07Added 6 answers

Multiply both sides of the inequality by $2pq$ to get $2q+2p>pq$, which is equivalent to $(p-2)(q-2)<4$. Now $p-2$ and $q-2$ are integers $\ge 1$, so it is easy to list the five possibilities:

$(p-2,q-2)\in \{(1,1),(1,2),(1,3),(2,1),(3,1)\},$

which implies that

$(p,q)\in \{(3,3),(3,4),(3,5),(4,3),(5,3)\}.$

$(p-2,q-2)\in \{(1,1),(1,2),(1,3),(2,1),(3,1)\},$

which implies that

$(p,q)\in \{(3,3),(3,4),(3,5),(4,3),(5,3)\}.$

Most Popular Questions