Proving 1>0: Why Does the Contradiction Method Fail?

[mrroboto]

1>0

how do you prove this?

i tried by contradiction but it doesn't work

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

but this doesn't work since -1>-2 does not imply that 1>4.

 

[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.

 

[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

 

[tiny-tim]

1>0

how do you prove this?

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.

 

[ice109]

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.

obviously using the usual ordering it's a theorem.

 

[HallsofIvy]

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.

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

I think 1 > 0 is an axiom.

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.

 

[cowah22]

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

 

[ice109]

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

(-1)^3 = 1?

 

[cowah22]

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

 

[quantumdude]

I don't think you can prove it.

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.

 

[K.J.Healey]

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.

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

 

[Petter]

(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

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. #

 

[Petter]

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

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

 

[HallsofIvy]

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".

 

[tiny-tim]

No it is not an axiom. The axioms for order (on the real numbers) are: …
3) if a< b and c> 0 then ac< bc.
…

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.

 

[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.

 

[Gokul43201]

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?

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

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).

 

[ice109]

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

ordering relations aren't preserved under even exponentiation

 

[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

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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.

 

[abelian jeff]

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

 

[cowah22]

ordering relations aren't preserved under even exponentiation

why?

 

[robert Ihnot]

I think a problem is that it was never decided--or at first even considered--What are the Axioms we are Employing? If it is to be assumed that we "know" what the positive integers are, then there is still the problem of what negative integers are. Suppose a<b and c=0, then is ac<bc?

I think what is needed is also the law of distribution (a+b)c=ac+bc. And by communitivity, this must also be the same as c(a+b). Generally at some point it must be shown or axiomatic that a+0 = 0+a = a, also ax0 = 0. And with negative numbers we have a+-a = 0. But for negative numbers the last equation can be the definition.

 

[abelian jeff]

why?

$-2 < 0$

but

$(-2)^2 =4 > 0^2=0$