# Thread: inequality proof

1. ## inequality proof

How do we prove : $\frac{1}{a}>0\Longrightarrow a>0$.

Proof:

Let $\frac{1}{a}>0$ and suppose $\neg(a>0)$,then by trichotomy law we have :

a<0 or a-0

For a<0 => $\frac{1}{a}.a =1<0$ ,hence contradiction ,since 1>0

For a=0,how do we cope with this case??

2. ## Re: inequality proof

there is no $a=0$ case. Division by $0$ is undefined.

3. ## Re: inequality proof

Originally Posted by romsek
there is no $a=0$ case. Division by $0$ is undefined.
So there is no theorem : (1/a) >0 => a>0 ??

4. ## Re: inequality proof

What does "there is no a= 0 case" have to do with "if (1/a)> 0 then a> 0"? Why do you conclude that the theorem is not true?

5. ## Re: inequality proof

Originally Posted by HallsofIvy
What does "there is no a= 0 case" have to do with "if (1/a)> 0 then a> 0"? Why do you conclude that the theorem is not true?
If there is no a=0 case ,we cannot prove: (1/a)>0 => a>0.
I cannot prove the theorem .

Perhaps you can??

6. ## Re: inequality proof

Originally Posted by psolaki
If there is no a=0 case ,we cannot prove: (1/a)>0 => a>0.
I cannot prove the theorem .

Perhaps you can??
Given $\dfrac 1 a > 0$

suppose $a < 0$ then

$a = -|a|$

$\dfrac 1 a = \dfrac {1}{-|a|}=-\dfrac{1}{|a|}=-\left|\dfrac 1 a\right|$

by definition

$\left|\dfrac 1 a\right| > 0$

so

$-\left|\dfrac 1 a\right| < 0$

and thus

$\dfrac 1 a < 0$

but we're given that

$\dfrac 1 a > 0$

and thus we have a contradiction so it must be that

$a > 0$

7. ## Re: inequality proof

Originally Posted by psolaki
How do we prove : $\frac{1}{a}>0\Longrightarrow a>0$.

Proof:

Let $\frac{1}{a}>0$ and suppose $\neg(a>0)$,then by trichotomy law we have :

a<0 or a-0

For a<0 => $\frac{1}{a}.a =1<0$ ,hence contradiction ,since 1>0

For a=0,how do we cope with this case??
The product of two positive numbers (and so the quotient of two positive numbers) gives a positive number.

The product of two negative numbers (and so the quotient of two negative numbers) gives a negative number.

The product of two numbers of different sign (and so the quotient of two numbers of different sign) gives a negative number.

Since your quantity 1/a is positive, that means the two numbers must be of the same sign. Since 1 is positive, so must be a. Q.E.D.

8. ## Re: inequality proof

Originally Posted by romsek
Given $\dfrac 1 a > 0$

suppose $a < 0$ then

$a = -|a|$

$\dfrac 1 a = \dfrac {1}{-|a|}=-\dfrac{1}{|a|}=-\left|\dfrac 1 a\right|$

by definition

$\left|\dfrac 1 a\right| > 0$

so

$-\left|\dfrac 1 a\right| < 0$

and thus

$\dfrac 1 a < 0$

but we're given that

$\dfrac 1 a > 0$

and thus we have a contradiction so it must be that

$a > 0$

ΝΟ If you have a contradiction you must have:

$\neg(a<0)$ and not a>o

But $\neg(a<0)$ implies by trichotomy law : a>0 OR a=0 and not only a>0

9. ## Re: inequality proof

Originally Posted by Prove It
The product of two positive numbers (and so the quotient of two positive numbers) gives a positive number.

The product of two negative numbers (and so the quotient of two negative numbers) gives a negative number.

The product of two numbers of different sign (and so the quotient of two numbers of different sign) gives a negative number.

Since your quantity 1/a is positive, that means the two numbers must be of the same sign. Since 1 is positive, so must be a. Q.E.D.
The product of $\frac{1}{a}.a=1$ IF a is NOT zero

But we donot know that

10. ## Re: inequality proof

Yes, we do know that. The very fact that we are talking about $\frac{1}{a}$ tells us that a is not 0.

11. ## Re: inequality proof

Originally Posted by HallsofIvy
Yes, we do know that. The very fact that we are talking about $\frac{1}{a}$ tells us that a is not 0.
There is no theorem,axiom or definition that tell us:

if (1/a) exists then $\neg(a=0)$

The only axiom we have tell us what happens IF $\neg(a=0)$.

And : $\neg(a=0)\Longrightarrow \frac{1}{a}.a=1$

12. ## Re: inequality proof

You don't need it. If there exist x such that $x\left(\frac{1}{0}\right)= 1$ then $x= 1(0)= 0$. But 0 times any number is 0 so that 1/x exist only if "0" is the only number in your set (and most field axioms include the axiom that the set of "numbers" contains more than one number). In any field, 0 has no multiplicative inverse: the is NO $\frac{1}{0}$.

13. ## Re: inequality proof

Originally Posted by psolaki
There is no theorem,axiom or definition that tell us:

if (1/a) exists then $\neg(a=0)$

The only axiom we have tell us what happens IF $\neg(a=0)$.

And : $\neg(a=0)\Longrightarrow \frac{1}{a}.a=1$
What's your story anyway? Every thread you've started you end up arguing with people trying to answer your questions.

If you have all the answers already just go away.

14. ## Re: inequality proof

Originally Posted by romsek
What's your story anyway? Every thread you've started you end up arguing with people trying to answer your questions.

If you have all the answers already just go away.
If i had all the answers i would not ask in the 1st place,would i

And if somebody gives me wrong answers ,what must i do

15. ## Re: inequality proof

If you do not have the answers, how do you know you are being given wrong answers?

If you believe that an answer given is wrong (which certainly can happen!) ask for an explanation of clarification.