# 1>0 how do you prove this?

1. Apr 30, 2008

### mrroboto

1>0

how do you prove this?

i tried by contradiction but it doesnt work

suppose 1<0
then -1>0
then -1^2 > 0^2
then1>0
contradiction.

but this doesnt work since -1>-2 does not imply that 1>4.

2. Apr 30, 2008

### mutton

Right, so you've found that when you square both sides of an inequality, the sign doesn't always stay the same. (When does the sign change? When does it stay the same?)

Here's a proof by contradiction:

Suppose 1 $$\leq$$ 0. (Notice that I'm not using a < sign here. You left out the case that 1 = 0.)
But this is false. (Why?)
Therefore, 1 > 0.

3. Apr 30, 2008

### ice109

his proof is fine and there's no need to consider the equals case because 1!=0.
for squaring and the direction of the inequality one should go back to the definition :

-1>-2 => ( -1-(-2) )^2 > 0^2 => 1>0 but does not imply 1>4

4. Apr 30, 2008

### tiny-tim

Hi mrroboto!

I don't think you can prove it.

> is just an ordering, usually from left to right.

You can just as well order from right to left.

In other words: if you define > to be the opposite of what it usually is, then all the arithmetic operations still work, but 1 < 0, all squares are < 0, and so on.

I think 1 > 0 is an axiom.

5. Apr 30, 2008

### ice109

obviously using the usual ordering it's a theorem.

6. Apr 30, 2008

### HallsofIvy

Staff Emeritus
You can't prove anything if it has to work for all possible definitions!

No it is not an axiom. The axioms for order (on the real numbers) are:
1) a< b is a transitive relation: if a< b and b< c, then a< c.
2) If a< b and c is any real number then a+c< b+ c.
3) if a< b and c> 0 then ac< bc.
4) if a and b are any real numbers then one and only one of these is true:
i) a= b
ii) a< b
iii) b< a

Apply 4 to a= 1, b= 0
Obviously $1\ne 0$
if 1< 0, then, adding -1 to both sides, 0< -1. From that, and 3, multiplying both sides by -1, 0< (-1)(-1)= 1, a contradiction.

Leaving us with iii: 0< 1.

7. Apr 30, 2008

### cowah22

how about -1 < 1 .. 1 > -1 .. 1^3 > -1^3 .. 1 > 1

8. Apr 30, 2008

### ice109

(-1)^3 = 1???

9. Apr 30, 2008

### cowah22

try.. -1 < 1 .. 1 > -1 .. 1^2 > -1^2 .. 1 > 1

10. Apr 30, 2008

### Tom Mattson

Staff Emeritus
You sure can, it appeared on one of my undergraduate exams many moons ago! HallsofIvy listed the ordering axioms for an ordered field. 0<1 requires a proof, since it isn't listed there.

11. Apr 30, 2008

### K.J.Healey

Why? But from "3" c must be greater than 0. And you let c= -1?

12. Apr 30, 2008

### Petter

Assume 1 < 0. Then by (2) 0 < -1. This gives us, using (3), that -1 < 0. Using (2) again we get 0 < 1, a contradiction. Since 0!=1 we are, using (4), done. #

13. Apr 30, 2008

### Petter

He previously proved that 0 < -1, so that's correct.

14. Apr 30, 2008

### HallsofIvy

Staff Emeritus
As Petter pointed out, I was doing a "proof by contradiction". If 1< 0, then 0< -1[/b], so we can say 0= 0(-1)< (-1)(-1)= 1 which contradicts "If 1< 0".

15. Apr 30, 2008

### tiny-tim

But if we replace axiom 3 with 1 > 0, then we can prove "if a< b and c> 0 then ac < bc" … and if we replace axiom 3 with 1 < 0, then we can prove "if a< b and c> 0 then ac > bc".

I can do this quite easily for the rational numbers:
Let's define the affirmative whole numbers to be 1, 2, 3, …, where 2 = 1 + 1, 3 = 1 + 1 + 1, … (this is the standard Peano definition, isn't it?);

if 1 > 0 then 1 + 1 > 1 + 0 = 1 > 0, and generally any affirmative whole number is > 0;

and if 1 < 0 then any affirmative whole number is < 0;

And let's define the affirmative simple fractions to be 1/2, 1/3, …, where 1/2 + 1/2 = 1, 1/3 + 1/3 + 1/3 = 1, …;

again, these are all < 0 or all < 0 according to whether 1 < 0 or 1 > 0.

And let's define the affirmative rationals to be any product of an affirmative whole number and an affirmative simple fraction; in other words, n/m, where both n and m are affirmative whole numbers;

and if 1 > 0, then all affirmative rationals are > 0; and if 1 < 0, then all affirmative rationals are < 0.

And let's define the rejective whole numbers and simple fractions and rationals in the same way, but using -1 instead of 1;

then if 1 > 0, then all rejective rationals are < 0; and if 1 < 0, then all rejective rationals are > 0.

By definition, an affirmative rational times a rejective rational is a rejective rational (because it is an affirmative multiple of -1).

(And incidentally a rejective rational times a rejective rational is an affirmative rational (because it is an affirmative multiple of -1.-1, which equals 1 because 0 = (1 + -1)(1 + -1) = 1 + -1 + -1 + -1.-1).)

And any rational is either affirmative or rejective or zero.

So if a b and c are rationals: if a< b and c > 0,

then a-b < 0, so one of a-b and c is affirmative, and the other is rejective, and so their product (a-b)c is rejective.

So if 1 > 0, then (a-b)c < 0, and so ac < bc;
but if 1 < 0, then (a-b)c > 0, and so ac > bc.

16. Apr 30, 2008

### robert Ihnot

1>0 how do you prove this?

There is a problem with 1 is not equal to 0, since orginally it was shown the postulates were consistant if 1 =0.

Now with the axioms in flux or something, since I thought Peano had only 4? axioms to start with, I will refer to Wikipedia which lists 9 axioms.

Axiom 7 tell us: For every natural number n, S(n) ≠ 0. That is, there is no natural number whose successor is 0.

Courant was very skeptical of certain elementary proofs such as (-1)(-1) = 1 since he believed that such things were fundamental to the system.

17. Apr 30, 2008

### Gokul43201

Staff Emeritus
But doesn't this contradict axiom 7 listed below?

So we can not have 1=0 since 1=S(0), by construction, and we can not have S(0) = 0 (using symmetry of equality).

18. Apr 30, 2008

### ice109

ordering relations aren't preserved under even exponentiation

19. Apr 30, 2008

### robert Ihnot

Gokul43201 :Originally Posted by robert Ihnot
1>0 how do you prove this?

There is a problem with 1 is not equal to 0, since orginally it was shown the postulates were consistant if 1 =0.

But doesn't this contradict axiom 7 listed below?

Wait! I was originally meaning to quote morriboto, who started this problem and said
: 1>0

--------------------------------------------------------------------------------

1>0 how do you prove this?

Sure it contradicts Axiom 7, but that, I assume, was not the original wording of the Peano postulates. However, with an internet check, I can't find any reference to an earlier case of the axioms where 0 satisfied all requirements. Though, I was told that at one point Peano made that mistake in his axiomatic setup.

Last edited: Apr 30, 2008
20. Apr 30, 2008

### abelian jeff

We know that for all real numbers a, $a^2 \geq 0$. But $1^2 = 1$ and $1 \neq 0$, hence $1 > 0$.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?