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

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SUMMARY

The discussion centers on proving the statement "for all real numbers x, there exists a real number y such that x < y." The proposed solution is to set y = x + 1, which correctly demonstrates that x is less than y. This approach is valid and confirms the existence of such a y for any real number x, addressing the confusion raised by another participant regarding the correctness of this proof.

PREREQUISITES
  • Understanding of real numbers and their properties
  • Basic knowledge of mathematical inequalities
  • Familiarity with existential quantifiers in logic
  • Experience with elementary algebraic manipulation
NEXT STEPS
  • Study the properties of real numbers and their ordering
  • Learn about existential and universal quantifiers in mathematical logic
  • Explore proofs involving inequalities and their applications
  • Review basic algebraic concepts, particularly manipulation of equations
USEFUL FOR

Students of mathematics, particularly those studying real analysis or introductory proofs, as well as educators looking for examples of existential proofs in real number theory.

Andrax
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Homework Statement


prove that
([itex]\forall[/itex]x[itex]\in[/itex]R) ([itex]\exists[/itex]y[itex]\in[/itex]R) : 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 doesn't
 

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