- #1

- 35

- 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.

- Thread starter mrroboto
- Start date

- #1

- 35

- 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

- 179

- 0

Here's a proof by contradiction:

Suppose 1 [tex]\leq[/tex] 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

- 1,707

- 5

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

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Hi mrroboto!1>0

how do you prove this?

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

- 1,707

- 5

obviously using the usual ordering it's a theorem.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.

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 962

You can't prove anything if it has to work forHi 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.

No it is not an axiom. The axioms for order (on the real numbers) are:I think 1 > 0 is an axiom.

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 [itex]1\ne 0[/itex]

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

- 15

- 0

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

- #8

- 1,707

- 5

(-1)^3 = 1???how about -1 < 1 .. 1 > -1 .. 1^3 > -1^3 .. 1 > 1

- #9

- 15

- 0

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

- #10

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,500

- 8

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.I don't think you can prove it.

- #11

- 609

- 0

Why? But from "3" c must be greater than 0. And you let c= -1?You can't prove anything if it has to work forallpossible 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 [itex]1\ne 0[/itex]

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.

- #12

- 6

- 0

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. #(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

- #13

- 6

- 0

He previously proved that 0 < -1, so that's correct.Why? But from "3" c must be greater than 0. And you let c= -1?

- #14

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 962

- #15

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

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".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.

…

I can do this quite easily for the rational numbers:

Let's define the

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

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.

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

- 1,056

- 0

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:

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

Gokul43201

Staff Emeritus

Science Advisor

Gold Member

- 7,051

- 18

But doesn't this contradict axiom 7 listed below?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.

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

- #18

- 1,707

- 5

ordering relations aren't preserved under even exponentiationtry.. -1 < 1 .. 1 > -1 .. 1^2 > -1^2 .. 1 > 1

- #19

- 1,056

- 0

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:

- #20

- 32

- 0

- #21

- 15

- 0

why?ordering relations aren't preserved under even exponentiation

- #22

- 1,056

- 0

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.

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.

Last edited:

- #23

- 32

- 0