Prove that for all real numbers x there is a y n which case x<y

Andrax
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Homework Statement


prove that
(\forallx\inR) (\existsy\inR) : x < y


Homework Equations





The Attempt at a Solution


x < y
y= x+1
then x<x+1
which is correct but I'm kinda not sure of this answer...
 
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y=x+1 satisfies x<y so what wrong with it?
 
hedipaldi said:
y=x+1 satisfies x<y so what wrong with it?

just wanted to make sure someone actually confused me saying it dosen't
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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