Proving the Existence of b: B-->A for Tricky Algebra Proof

  • Thread starter Thread starter Pearce_09
  • Start date Start date
  • Tags Tags
    Algebra Proof
Click For Summary
SUMMARY

The discussion centers on proving the existence of a mapping \( b: B \to A \) such that \( bR = id_A \), where \( R: A \to B \) is a given mapping. The participants explore the implications of the condition \( RT = RS \) leading to \( T = S \) and consider the injectivity of \( R \) as a potential approach to the proof. The key focus is on understanding the relationship between the mappings and the identity function in the context of algebraic structures.

PREREQUISITES
  • Understanding of algebraic mappings and functions
  • Familiarity with concepts of injectivity and identity functions
  • Knowledge of the properties of function composition
  • Basic principles of algebraic proofs and logic
NEXT STEPS
  • Study the properties of injective functions in algebra
  • Learn about the concept of identity mappings in algebraic structures
  • Explore the implications of function composition in proofs
  • Investigate the role of left cancellability in algebraic proofs
USEFUL FOR

Mathematicians, algebra students, and educators interested in advanced algebraic proofs and the properties of mappings between sets.

Pearce_09
Messages
71
Reaction score
0
tricky algebra proof...

hello,

consider the mappings: (R: A-->B)

Suppose T: C-->A and S: C-->A satisfy RT = RS then T = S
prove that there exists a; b: B-->A such that bR = idA (identity of A)

well, I am not sure if I can say that b (inverse) = R
since b maps B to A .. and R maps A to B... and if i can say that...
how do i approach the proof?

I know that RT = RS then T = S.. should i work with this.? using b (inverse)...
or should i try to see if R or b is injective?...

ie. b(inverse)T = b(inverse)S ... ??
 
Physics news on Phys.org
The idea of left cancellable (RS=RT => S=T) is exactly the same as R being injective.
 

Similar threads

Direct Proof that every zero of p(T) is an eigenvalue of T
  • · Replies 24 ·
Replies
24
Views
4K
Prove ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates
  • · Replies 3 ·
Replies
3
Views
2K
Prove ##a\cdot b = b \cdot a ##using Peano postulates
  • · Replies 3 ·
Replies
3
Views
1K
Prove if ##a\cdot c = b \cdot c## then ##a = b## using Peano postulates
  • · Replies 2 ·
Replies
2
Views
1K
Prove ##1 + a=s(a)=a+1## for ##a \in \mathbb{N}##
  • · Replies 1 ·
Replies
1
Views
2K
Prove ##a + b = b + a## using Peano postulates
  • · Replies 3 ·
Replies
3
Views
2K
Prove ##a \cdot 1 = a = 1 \cdot a## for ##a \in \mathbb{N}##
  • · Replies 1 ·
Replies
1
Views
2K
Prove ##(a\cdot b)\cdot c =a\cdot (b \cdot c)## using Peano postulates
  • · Replies 1 ·
Replies
1
Views
1K
Hard MVT theorem proof Tutorial Q7
  • · Replies 12 ·
Replies
12
Views
2K
Prove ##(a+b) + c = a + (b+c)## using Peano postulates
  • · Replies 9 ·
Replies
9
Views
2K