1>0 how do you prove this?

  • Thread starter mrroboto
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In summary: I agree that 0<1 is an axiom.The axiom I was referring to is 3) above, which is used in the "proof" 1 > 0.
  • #1
35
0
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.
 
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  • #2
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 [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
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
mrroboto said:
1>0

how do you prove this?

Hi mrroboto! :smile:

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. :smile:

I think 1 > 0 is an axiom.
 
  • #5
tiny-tim said:
Hi mrroboto! :smile:

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. :smile:

I think 1 > 0 is an axiom.

obviously using the usual ordering it's a theorem.
 
  • #6
tiny-tim said:
Hi mrroboto! :smile:

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. :smile:

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 [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
how about -1 < 1 .. 1 > -1 .. 1^3 > -1^3 .. 1 > 1
 
  • #8
cowah22 said:
how about -1 < 1 .. 1 > -1 .. 1^3 > -1^3 .. 1 > 1

(-1)^3 = 1?
 
  • #9
try.. -1 < 1 .. 1 > -1 .. 1^2 > -1^2 .. 1 > 1
 
  • #10
tiny-tim said:
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.
 
  • #11
HallsofIvy said:
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 [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.

Why? But from "3" c must be greater than 0. And you let c= -1?
 
  • #12
(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. #
 
  • #13
K.J.Healey said:
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.
 
  • #14
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
HallsofIvy said:
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.
 
  • #16
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
robert Ihnot said:
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).
 
  • #18
cowah22 said:
try.. -1 < 1 .. 1 > -1 .. 1^2 > -1^2 .. 1 > 1

ordering relations aren't preserved under even exponentiation
 
  • #19
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.
 
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  • #20
We know that for all real numbers a, [itex]a^2 \geq 0[/itex]. But [itex]1^2 = 1[/itex] and [itex]1 \neq 0[/itex], hence [itex]1 > 0[/itex].
 
  • #21
ice109 said:
ordering relations aren't preserved under even exponentiation

why?
 
  • #22
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.
 
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  • #23
cowah22 said:
why?

[itex]-2 < 0[/itex]

but

[itex](-2)^2 =4 > 0^2=0[/itex]
 

1. What does "1>0" mean?

The notation "1>0" means "1 is greater than 0". This is a mathematical statement that is read as "one is larger than zero".

2. Why is it important to prove this statement?

Proving "1>0" is important because it is a fundamental concept in mathematics. It helps us understand the relationship between numbers and is the basis for many other mathematical principles.

3. How can you prove that "1>0" is true?

This statement can be proven using the properties of real numbers. We know that the number 1 is greater than 0 because it lies to the right of 0 on the number line. Additionally, when comparing two positive numbers, the one with a larger value is always greater. Therefore, 1>0 is a proven and accepted mathematical fact.

4. Can you give an example to show that "1>0" is true?

Yes, one example would be to take any two positive numbers, such as 5 and 3. We know that 5 is larger than 3, so we can say that 5>3. Since both 5 and 3 are greater than 0, we can also say that 5>0 and 3>0. This means that 5 is greater than both 0 and 3, therefore 5>0 and 3>0 are both true statements. And since 5>0 and 5>3, we can conclude that 5>3>0, or 1>0.

5. Are there any real-world applications of "1>0"?

Yes, the concept of "1>0" is used in many real-world applications, including engineering, finance, and statistics. For example, in engineering, the concept of "1>0" is used to determine the strength and stability of structures. In finance, it is used to compare interest rates and investments. In statistics, it is used to determine the likelihood of an event occurring. Overall, understanding and proving "1>0" is crucial in many fields of study.

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