Question

Prove that if x=2, x != 1. @jim_thompson5910 @ganeshie8 @AravindG @phi @robtobey @Compassionate @eliassaab @Vincent-Lyon.Fr @skullpatrol @Kainui @kc_kennylau @inkyvoyd @Andras @tyteen4a03 @divu.mkr @arabpride @wolfe8 @Yttrium @Kristen17 @christianfilms11 @Owlcoffee @lauluisa @arig1221 @INeedHelpPlease? @OpenSessame @Confusionist @lucaz @Figureskater120 @Kittta @sweetburger @ShaeLoveth♥ @Hawaiian_life @IDAREYOU @leozap1 @JMark @harpreetsk @CrayolaCrayon_ @river_song @SWAG2.0 @link,zoro @Kaylalove22 @zubhanwc3 @Cutefriendzoned @HarukaN @christianfilms11 @christopher_miller420 @ayeshaafzal221 @st

Answer 1

If you mass tag again

Answer 2

o-o wow xD this kind of makes me not wanna help :\

Answer 3

We have \(1\ne2\).
QED

Answer 4

@kc_kennylau My proefssor says your reasoning doesn't say anything related to the problem.

Answer 5

Attention: != means \(\ne\)

Answer 6

@FutureMathProfessor Who the hell is your professor, is it yourself

Answer 7

@kc_kennylau My professor says your reasoning isn't enough she said you need to show explicitly that 1 != 2 and then that X can not occupy both values at the same time.

Answer 8

FMP, that's the correct way to do the problem. What's slipping you up? Does your professor not 'like' the way we're doing this, or is it just confusing you and you're embarrassed? Also, we've asked you not to mass tag multiple times.

Answer 9

this is advanced number theory; 2 != 1 is not sufficient. We need to prove it.
@ikram002p have a look when you have time

Answer 10

@FutureMathProfessor , these questions are exceedingly annoying when you mass tag. Not only that, but you haven't given us a set of axioms to prove with, so your question is moot.

Answer 11

lol

Answer 12

ikram help me on my question

Answer 13

the proof depand on an axiom system u cud have , let assume we have this
familiar axiom sys
{ let 0,a,b belong to z
a+0=a
a+-a=0
....... let s
={ for every n belong to Z , n+1 >n }
or oly use axiom of 0 <1 }
use contradicton
let x= 1 , and x= 2
x-1=0
x-2=0
x-1=x-2
-1=-2
1=2
0=1 which is a contradiction to 0<1