proof 0 < 1

luca-deltodesco · 2007-02-02 11:00:19

a friend gave me this as a little challenge, would this be a correct way of proving this?

Given two integers a,b there are only 3 things we can say about them a<b, a = b, a>b

Assuming that 1 < 0, subtracting 1 from both sides gives
0 < -1, multiplying both sides by -1, gives
0 < 1, however this is a contradiction, as this would mean that both statements 1 < 0, 0 < 1 are true, but 1 cannot be both less than , and greater than 0, therefore 1 < 0 cannot be true, since 0 obviously doesnt equal 1, this leaves us with only one alternative, 0 < 1

Ricky · 2007-02-02 11:07:58

Nope. When you divide or multiply by a negative, the inequality sign "<" must flip.

I don't believe it is possible to prove this without using some rigorous definition somewhere. For example, one may prove that the multiplicative identity is the smallest positive number in a ring with unity, if there is a smallest positive integer. This would of course mean that 0 < 1 as 1 - 0 = 1 and 1 is in the positive numbers.

Or you may in turn define integers by pairs of natural numbers [a, b] which, naively, mean "a - b". So [n, n] acts as 0, [n+1, n] acts as 1. Then the proof becomes simple once you define "<" to mean that for any pair [a, b] and [c, d], that [a, b] < [c, d] if and only if a + d < c + b.

luca-deltodesco · 2007-02-02 11:55:04

because i made a mistake

MathsIsFun · 2007-02-02 12:00:32

Got us thinking, though!

Math Is Fun Forum

#1 2007-02-02 11:00:19

proof 0 < 1

#2 2007-02-02 11:07:58

Re: proof 0 < 1

#3 2007-02-02 11:55:04

Re: proof 0 < 1

#4 2007-02-02 12:00:32

Re: proof 0 < 1

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