# prove a=a (natural numbers)Watch

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#1
I'm trying to prove that a=a if a is a natural number by contradiction suggesting that if a<a then a=a+x where x is a natural number. However in doing so I go on to use the additive cancellation property of the addition of natural numbers and suggest that _ = x. Is this sufficient? I don't want to introduce 0 as I would then be transcending the stipulations of the natural numbers to prove something of the natural numbers which I feel undermines the rigor of the proof. How do I show ' nothing'? Or can I use 0 and then go on to say that 0 is not a natural number and prove that 1 is the least natural number thus 0 is not and consequently a<a and a>a is untrue so a=a must be true?
0
5 years ago
#2
seriously, thats what you do on Friday evening?
2
5 years ago
#3
a stands for 1 number no? Then I don't think you can say a<a . Or maybe I am just stupid
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#4
(Original post by Aniaa)
a stands for 1 number no? Then I don't think you can say a<a . Or maybe I am just stupid
But there's no axiom to directly suggest that a=a
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#5
(Original post by Aniaa)
seriously, thats what you do on Friday evening?
Sorry but what are YOU doing? Lol... Commenting on what I'm doing? Sounds like you're having an epic Friday night.
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5 years ago
#6
(Original post by Jam')
Sorry but what are YOU doing? Lol... Commenting on what I'm doing? Sounds like you're having an epic Friday night.
But you're commenting on someone commenting on what you're doing so inturn your Friday night is just as bad?
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5 years ago
#7
(Original post by Jam')
But there's no axiom to directly suggest that a=a
What axioms do you have?
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5 years ago
#8
First of all by the semantics assigned to '=', by default a = a is always valid. The proof is trivial, you can just use proof by induction.
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