Is my proof correct? (show that if a < 0, then -a > 0)

[Mépris]

1. Homework Statement :

Show that if a < 0, hen -a > 0

Homework Equations

N/A

3. The Attempt at a Solution :

Assume the result is false.

Thus, by trichotomy:

i)

-a = 0

This is not possible. A real number can be greater than zero, less than zero or equal to zero.
Since it has been given that -a > 0, it follows that -a ≠ 0

ii)

-a < 0

If -a < 0, then a > 0

a + (-a) < 0 + a

\Rightarrow a > 0, which is not true!

Since, it is given that a < 0

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This is my first foray into learning calculus in a more rigorous fashion. I've never done any proofs before. This seems okay to me but it is possible that there is something wrong somewhere. So, is this indeed correct or did I miss something?

Thank you

 

[Infinitum]

Hi Mépris!

-a = 0

This is not possible. A real number can be greater than zero, less than zero or equal to zero.
Since it has been given that -a > 0, it follows that -a ≠ 0

You aren't given that fact. That's what you are trying to prove..

ii)

-a < 0

If -a < 0, then a > 0

a + (-a) < 0 + a

⇒ a > 0, which is not true!

Since, it is given that a < 0

This part sounds correct!

 

[Mépris]

Hello Inifinitum!

You aren't given that fact. That's what you are trying to prove..

/facepalm

This is actually a two-part question but I didn't think posting the first part would be necessary. Here is the full question:
"Show that if a > 0, then -a < 0; and if a < 0, then -a < 0"

The solution to the first part was provided and it was said that the second part of the solution could be inferred from the first. So, I tried to prove it like the first part was proven and I didn't look closely enough. (complacency, I guess?)

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If anyone's wondering, the question is from Ian Craw's notes on "Advanced Calculus and Analysis", which is a first-year course in the math degree at the University of Aberdeen. I decided to start off my study of calculus with that because I found Courant (there is a free online copy on the AIT's website) a little too hard for my level and Craw's notes/book are written with people who've studied the A-Levels and Scottish Advanced Highers in mind. Since, I did A-Level Mathematics, these notes are a nice transition. :-) :-)

This part sounds correct!

BOOYA!

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Assume the result is false.

Thus, by trichotomy:

i)

-a = 0

Multiplying by (-1) on both sides,

a x (-1) = 0 x (-1)
⇒ a = 0

It is given that a < 0,
so, it follows that 0 should be less than 0, since a = 0

This is not possible, because by trichotomy, a non-zero real number can be either equal to zero, greater than zero or less than zero. Zero cannot be less than zero!

The above is my second attempt, which looks correct. Nevertheless, I'm not certain about my wording. Is the way I write precise enough? Further, am I correct in saying that bit about a "non-zero real number..."? I ask because it was not explicitly stated in the definition given in the notes. After the definition, it is stated that "a real number is either positive, negative or zero". Can I just use that when talking about "trichotomy"?

Thank you.

 

[Infinitum]

by trichotomy, a non-zero real number can be either equal to zero, greater than zero or less than zero. Zero cannot be less than zero!

Further, am I correct in saying that bit about a "non-zero real number..."?

Nooo! How can a non-zero real number be equal to zero!?

When you got, a=0, you could immediately say that since its already given,a &lt; 0, and by identity 0=0, hence-a \neq 0, causing a contradiction.

 

[Mépris]

Nooo! How can a non-zero real number be equal to zero!?

When you got, a=0, you could immediately say that since its already given,a &lt; 0, and by identity 0=0, hence-a \neq 0, causing a contradiction.

Okay, that was really stupid of me - tried to rush through it and just added that there. I can actually feel my cheeks flushing, which I don't think I've ever felt before. :$

Great. Thanks a lot!