Proofing Calculus 1: Prove (-1)*a=-a

[SarRis]

Homework Statement

Prove that (-1)*a=-a

Relevant Equations

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Hello! So I just started calculus 1 at my university and I have a problem with proofs since I haven't done them ever before. It would be great if you could help me with this question and give me a few advice on how to approach proofs (books/your personal tips/whatever you remember).
Thank you!

 

[berkeman]

Welcome to PF.

What have you covered in class so far? Do you have some axioms that you can use for this proof? You must have been taught something that can be applied here, no?

 

[PeroK]

According to the homework guidelines you have to make a good attempt at this yourself.

For a proof like this you must rely on how things are defined precisely. In anything but a rigorous algebra course, it would be taken for granted that $(-1)a = -a$. So, you are going to have to figure out what those two sides of the equation actually mean before you can show that they are equal.

 

[SarRis]

So before this problem, we had two lessons. Defining the set of real numbers (axiomatic aproach). We defined binary oprations +,-, *, : and binary relations <, >, <=, >=. Also commutativity, associativity, identity element, inverse element. (I hope I explained well, English is not my mother tongue)
Why I did not write out my attempt at this problem is because I am just starting with proofs and I actually don't know how to approach them at all.
Thank you for understanding.

 

[PeroK]

So before this problem, we had two lessons. Defining the set of real numbers (axiomatic aproach). We defined binary oprations +,-, *, : and binary relations <, >, <=, >=. Also commutativity, associativity, identity element, inverse element. (I hope I explained well, English is not my mother tongue)
Why I did not write out my attempt at this problem is because I am just starting with proofs and I actually don't know how to approach them at all.
Thank you for understanding.

What is $-a$? How would you define/describe that? And don't say "negative $a$"!

 

[SarRis]

What is $-a$? How would you define/describe that? And don't say "negative $a$"!

So based on what we did before. I would say that -a is the inverse element of a.
Maybe the approach is to prove that (-1)a is the inverse element of a.

 

[PeroK]

So based on what we did before. I would say that -a is the inverse element of a.
Maybe the approach is to prove that (-1)a is the inverse element of a.

It's better to say the additive inverse of $a$. As opposed to the multiplicative inverse. And, yes, that's the right approach. What happens if you calculate $a + (-1)a$?

 

[SarRis]

It's better to say the additive inverse of $a$. As opposed to the multiplicative inverse. And, yes, that's the right approach. What happens if you calculate $a + (-1)a$?

You get the (additive?) identity/ neutral element. Zero 0.

 

[PeroK]

You get the (additive?) identity/ neutral element. Zero 0.

Yes, but that has to be proved step by step. Welcome to pure maths!

Hint: what law relates to both addition and multiplication?

 

[SarRis]

Thank you!
The law that relates to both addition and multiplication is commutativity.
So is it, just by adding the (-1)a and a, and getting the neutral element 0. We prove that (-1)a is the inverse element of a? Like this.
(-1)a + a = a + (-1)a = 0

 

[PeroK]

Thank you!
The law that relates to both addition and multiplication is commutativity.
So is it, just by adding the (-1)a and a, and getting the neutral element 0. We prove that (-1)a is the inverse element of a? Like this.
(-1)a + a = a + (-1)a = 0

Not quite. I meant the distributive law. I took four steps to show that $(-1)a + a = 0$.

And, I used two results which I assume you have covered: that $a.0 = 0$ and the uniqueness of the additive inverse.

 

[SarRis]

We did cover that a*0=0 and the uniqueness of the additive inverse.
Is it (-1)a + a = 0
a(-1 + 1) = 0
a*0=0
We actually didn't cover the distributive law before this. Although I know that a*(b+c)=a*b + a*c (from left and right both)

 

[SarRis]

No we did! Sorry! Oh now I see!

 

[SarRis]

Thank you very much! I had a small eureka moment :)

 

[PeroK]

We did cover that a*0=0 and the uniqueness of the additive inverse.
Is it (-1)a + a = 0
a(-1 + 1) = 0
a*0=0
We actually didn't cover the distributive law before this. Although I know that a*(b+c)=a*b + a*c (from left and right both)

This is not quite right. You can't immediately equate $(-1) + a$ to $0$. Instead, what I did was:
$$(-1).a + a = (-1).a + 1.a = (-1 + 1).a = 0.a = 0$$
From which we conclude that $(-1).a = -a$. Note that you could prove this last step by adding $-a$ to both sides of the equation and using the associative law. That might be a good exercise.

 

[SarRis]

Thank you once again!

 

[WWGD]

Thank you!
The law that relates to both addition and multiplication is commutativity.
So is it, just by adding the (-1)a and a, and getting the neutral element 0. We prove that (-1)a is the inverse element of a? Like this.
(-1)a + a = a + (-1)a = 0

Consider the Distributive law of addition with respect to product.