Addition problem in Serge Lang Basic Mathematics

AI Thread Summary
The discussion revolves around the manipulation of the expression -(a - b) using established mathematical rules. The original poster attempts to prove that -(a - b) equals b - a but questions the validity of their approach, particularly regarding the justification for treating -b as +(-b). Feedback suggests that while the solution isn't incorrect, a more abstract understanding of negation could clarify the reasoning. The book's solution, which utilizes the properties of additive inverses, is recommended as a more straightforward method. Overall, the conversation emphasizes the importance of rigor in proofs and the need to adhere to formal definitions.
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Homework Statement
Show that -(a - b) = b - a
Relevant Equations
-(a - b) = b - a
Relevant Rules:
N5: -(a+b) = - a - b
N4: a = -(-a)
N2: a + (-a) = 0 and -a + a = 0

I tried just manipulating -(a - b) with the rules to get the answer:
-(a - b) = -(a + (-b))
With N5: = - a + (-(-b))
With N4: = - a + b
Commutativity: b - a

The provided solution in the book used N2 to prove it:
1741381664719.png


Is my solution valid? If not what is the problem with it? I would really appreciate any feedback because I'm new to thinking about proofs and have no idea what I'm doing.

I guess saying -(a - b) = -(a + (-b)) might be wrong because I can't find any rule to justify it. I assumed that - b and + (- b) were assumed to be the same from this section:
1741381903423.png


But maybe I can't use things from more casually written sections in proofs?

Thanks for reading!
 
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clone said:
Is my solution valid?
Looks fine to me.
clone said:
I guess saying -(a - b) = -(a + (-b)) might be wrong because I can't find any rule to justify it.
Seems to me that it's covered by N4 or else by the fact that subtracting a number is the same as adding the additive inverse of that number.
 
N4 has to be proven. You also used ##-a=(-1)\cdot (+a)## and the distributive law without mention:
$$
-(a-b)=(-1)\cdot (a+(-b))=(-1)\cdot a+ (-1)\cdot (-b)=-a+(-(-b))=-a+b
$$
 
clone said:
Homework Statement: Show that -(a - b) = b - a
Relevant Equations: -(a - b) = b - a

Relevant Rules:
N5: -(a+b) = - a - b
N4: a = -(-a)
N2: a + (-a) = 0 and -a + a = 0

I tried just manipulating -(a - b) with the rules to get the answer:
-(a - b) = -(a + (-b))
With N5: = - a + (-(-b))
With N4: = - a + b
Commutativity: b - a

The provided solution in the book used N2 to prove it:
View attachment 358234

Is my solution valid? If not what is the problem with it? I would really appreciate any feedback because I'm new to thinking about proofs and have no idea what I'm doing.

I guess saying -(a - b) = -(a + (-b)) might be wrong because I can't find any rule to justify it. I assumed that - b and + (- b) were assumed to be the same from this section:
View attachment 358236

But maybe I can't use things from more casually written sections in proofs?

Thanks for reading!
Your solution isn't wrong, but perhaps you could look at this problem differently. From an abstract point of view, what is ##-X##, where ##X## is anything? The abstract mathematical answer is that ##-X## is the unique thing that satisfies the equation ##X + (-X) = 0##.

Whereas, you are looking at the ##-## as more like an operator that acts on ##X##.

From that point of view, I very much prefer the book solution.
 
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Likes SammyS and clone
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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