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1>0 how do you prove this?

  1. Apr 30, 2008 #1
    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. jcsd
  3. Apr 30, 2008 #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.
     
  4. Apr 30, 2008 #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
     
  5. Apr 30, 2008 #4

    tiny-tim

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    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.
     
  6. Apr 30, 2008 #5
    obviously using the usual ordering it's a theorem.
     
  7. Apr 30, 2008 #6

    HallsofIvy

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    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.
     
  8. Apr 30, 2008 #7
    how about -1 < 1 .. 1 > -1 .. 1^3 > -1^3 .. 1 > 1
     
  9. Apr 30, 2008 #8
    (-1)^3 = 1???
     
  10. Apr 30, 2008 #9
    try.. -1 < 1 .. 1 > -1 .. 1^2 > -1^2 .. 1 > 1
     
  11. Apr 30, 2008 #10

    Tom Mattson

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    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.
     
  12. Apr 30, 2008 #11
    Why? But from "3" c must be greater than 0. And you let c= -1?
     
  13. Apr 30, 2008 #12
    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. #
     
  14. Apr 30, 2008 #13
    He previously proved that 0 < -1, so that's correct.
     
  15. Apr 30, 2008 #14

    HallsofIvy

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    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".
     
  16. Apr 30, 2008 #15

    tiny-tim

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    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.
     
  17. Apr 30, 2008 #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.
     
  18. Apr 30, 2008 #17

    Gokul43201

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    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).
     
  19. Apr 30, 2008 #18
    ordering relations aren't preserved under even exponentiation
     
  20. Apr 30, 2008 #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.
     
    Last edited: Apr 30, 2008
  21. Apr 30, 2008 #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].
     
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