Let R be a ring with identity, and let x = R. Then the equation (-1)x= -x is true in R.
Let R be a ring with identity, and let x = R. Then the equation (-1)x= -x is true in R.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question

Transcribed Image Text:Let R be a ring with identity, and let x = R. Then the equation (-1)x= -x is true in R.
Fill in the white boxes to make a proof of this fact. For each step in the proof, fill in the grey
box to indicate which ring axiom or other fact is used to derive that step.
Proof:
• By
• By
• By
By
• By
●
the multiplicative identity law
the additive identity law
we have
we have
we have
we have
we have (-1)x= -X.
(You should fill all the boxes, so if you came up with a proof that doesn't fill them all, try
including more intermediate steps. Please feel free to look up in the lecture notes what
"Proposition 5.11" is.)
uniqueness of additive inverses
Proposition 5.11
›
"
the distributive law
the additive commutative law
the additive inverse law
(-1+1)x= 0 (-1)x+ 1x=0|| 0((-x) + x) = 0 0x = 0
(-1)(-x) = 1x (-1)x+x=0
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