How to prove statement with quantifiers?

Question

How would I prove something like this:

$ \forall x >0, \exists y\gt 0,(xy = 1) $

I read it as "For all $x$ greater than $0$, there exists a $y$ greater than $0$, such that $xy=1$."

Is this correct?

I understand this is such an easy proof and I assume it is obviously true, but I don't really understand where to start it or how to prove it. I can see that I eventually need to show that $\ y=1/x $ but how do I show that? Should I do it by cases? Should I do it by contradiction?

Answer 1

No need to do this case by case.

All we need is to input any/every real positive number $x$, (for every $x\in \mathbb R$, such that $x>0$,)

we can find a $y> 0, y\in \mathbb R$, specifically $$(y> 0,\; \;y\in \mathbb R)\; \;y=\frac 1x$$ such that that $$xy =x\cdot \frac 1x = 1$$

Why we consider $y= \dfrac 1x \gt 0$?:

Given $x>0,x\in \mathbb R$, can you show that $y=\dfrac 1x$ ensures us that $y>0,$ and $y\in \mathbb R$?

Answer 2

Let's say we look at $x,y \in \mathbb{Q}$ for now.

Well, you assume that you have an arbitrary $x > 0$. So you are not allowed to use any "special" properties of this number except its positivity.

Now you need to find (at least) one $y > 0$ such that xy = 1. One way to do that is to note that $x$ can be written as $x = \frac{a}{b}$ with both $a,b \in \mathbb{N}$, i.e. $a,b > 0$. Then the $y$ you are looking for is given by $y = \frac{b}{a} = \frac{1}{x}$. This is again a positive number, and you can easily check that $xy = 1$.

So the idea is that you take a very general $x$ with no other properties, and then your task is to find an associated $y$ such that $xy = 1$. In finding $y$, you may, however, use what you know about $x$, as $x$ is fixed after you have "chosen" to consider one. Then if you look at a different $x$, if you have done the proof right, you should be able to find the corresponding $y$, which then completes the proof.

Hope that helps!

Answer 3

A common structure for a proof of a 'for all...there exists' statement is to start by saying: Let $x$ be an arbitrary... and then producing an object that satisfies the conditions.

So in your problem, I would say: Let $x$ be an arbitrary positive real number. We can then take $y=\frac1x.$

Then $xy=x\cdot \frac1x=1$, as desired.

Answer 4

hint

When the statement begins by $$\forall x\in E $$ the proof should begin by

Let $x $ be an element of $E $.

When the statement contains $$\exists y\in B $$

there are two possibilities :

1.

the existence of $y $ is insured by a theorem (IVT, MVT, Rolle...)

2.

You construct your $y $ by

Put $y=... $.

In your case, we will say : Let $x $ be a real $>0$.

put $y=\frac {1}{x} $ then $xy=1$.

Answer 5

As the OP states, the question looks so simple that coming up with a proof looks daunting. But this is an opportunity to set up an interesting 'premise framework' and to proceed with a demonstration that builds on these axioms.

(P1) Every nonzero real number $x$ has a multiplicative inverse $y$ that is also not equal to zero.

(P2) The 'take the sign' mapping $\pi$ on the set of nonzero real numbers, defined by
$\quad \pi(x) = 1 \text{ for } x \gt 0$
and
$\quad \pi(x) = -1 \text{ for } x \lt 0$
satisfies $\pi(xy) = \pi(x) \pi(y)$ for all nonzero real numbers $x$ and $y$
(the product of two nonzero numbers can't be equal to zero).

Proposition: $\forall x >0, \exists y\gt 0,(xy = 1)$
Proof
Let $x$ be a positive real number. By (P1) $x$ has a nonzero inverse $y$. By (P2) we can apply $\pi$,

$\quad \pi(xy) = \pi(1) = 1 = \pi(x) \pi(y) = 1 \pi(y) = \pi(y)$

so we conclude that $\pi(y) = 1$. But the inverse of $\{1\}$ under $\pi$ is precisely the set of positive numbers, so $y$ must be positive. $\quad \blacksquare$

Answer 6

Although the following in not how proofs are normally done in practice (in most math textbooks etc.), I just wanted to show how it could be proven with the rules of natural deduction and algebraic axioms:

\begin{array}{l} & \{1\} & 1. & a \in \mathbb{R} \land a>0 & \text{ Assum. }\\ & - & 2. & a(\frac{1}{a})=a(\frac{1}{a}) & \text{ =I }\\ & \{1\} & 3. & a(\frac{1}{a})=1 & \text{ 1,2 Inverse (if $a \neq 0$) }\\ & - & 4. & 1>0 & \text{ Def. 1 }\\ & \{1\} & 5. & \frac{1}{a} > \frac{0}{a} & \text{ 1,4 Division (if $a > 0$) }\\ & \{1\} & 6. & \frac{1}{a} > 0 & \text{ 5 Mult. property of 0 }\\ & \{1\} & 7. & \frac{1}{a} \in \mathbb{R} & \text{ 1,5 Closure on $\mathbb{R}$ }\\ & \{1\} & 8. & \frac{1}{a} \in \mathbb{R} \land \frac{1}{a}>0 \land a(\frac{1}{a})=1 & \text{ 3,6,7 $\land$I }\\ & \{1\} & 9. & \exists y[y \in \mathbb{R} \land y>0 \land ay=1] & \text{ 8 EI }\\ & - & 10. & (a \in \mathbb{R} \land a>0) \implies \exists y[y \in \mathbb{R} \land y>0 \land ay=1] & \text{ 1,9 CP }\\ & - & 11. & \forall x[(x \in \mathbb{R} \land x>0) \implies \exists y[y \in \mathbb{R} \land y>0 \land xy=1]] & \text{ 10 UI }\\ \end{array}