Square Of A Number Is Non-negative

[Bashyboy]

Hello, I am seeking some aid in proving that the square of a number is always non-negative. Here is some of my proof:

A number, call it a, is either positive, negative, or zero. A number squared is produced when you take the number and multiply it by itself. So, we have three cases to consider.

When a < 0:

If a < 0, then 0 - a > 0; and so, 0 - a = -a must be in the set of positive numbers. If -a is in the set of positive numbers, then, by property P12 (Closure under multiplication: If a and b are in P, then a • b is in P), (-a)(-a) must be in the set of positive numbers, implying that it is a positive number, and the negative signs may vanish. (-a)(-a) = a x a = a^2 > 0.

Initially, I thought that this proof would be valid; but as I went along further along, I developed an inkling of a feeling that it was not so.

What is wrong with the proof of the case a > 0?

 

[HallsofIvy]

The only thing "wrong with the proof of the case a> 0" is that the "proof" says nothing about a> 0! You say "when a< 0" but give no other case.

Of course, if a> 0 then "property P12" itself immediately tells you that a*a is positive.

 

[Bashyboy]

No, I intentionally did not include the case when a > 0, because I had ran into trouble with the case when a < 0. Just to be clear, my proof for the case a < 0 is valid?

 

[Ray Vickson]

No, I intentionally did not include the case when a > 0, because I had ran into trouble with the case when a < 0. Just to be clear, my proof for the case a < 0 is valid?

That depends on what properties you are permitted to use. You say that (-a)*(-a) = a^2 (true!), but have you already been given this, or is this yet again another thing that needs to be proved first?

 

[phinds]

A number, call it a, is either positive, negative, or zero.

OR ... imaginary. Are you leaving those out on purpose? You didn't say so.

 

[Bashyboy]

The difficulty I am having with my proof is when I state, "If a < 0, then 0 - a > 0; and so, 0 - a = -a must be in the set of positive numbers." Doesn't that mean that -a is a positive number, and clearly not a negative number, implying the next step I proceed is simply the multiplication of two POSITIVE numbers, (-a)(-a), which doesn't prove anything about the multiplication of two NEGATIVE numbers? Or am I wrong, and my initial proof is correct?

 

[rcgldr]

I'm not sure what mathematical rules you're allowed for the proof, but perhaps these statements could be used:

(+1)(a) = a
(-1)(a) = -a
(+1)(+1) = 1
(-1)(-1) = 1
((a)(b))(c) = (a)((b)(c)) ; multiplication is associative

 

[Ray Vickson]

The difficulty I am having with my proof is when I state, "If a < 0, then 0 - a > 0; and so, 0 - a = -a must be in the set of positive numbers." Doesn't that mean that -a is a positive number, and clearly not a negative number, implying the next step I proceed is simply the multiplication of two POSITIVE numbers, (-a)(-a), which doesn't prove anything about the multiplication of two NEGATIVE numbers? Or am I wrong, and my initial proof is correct?

No, it proves exactly what you want. It all hinges on whether or not you know or are allowed to use the property a*a = (-a)*(-a). Remember me asking about that in my first response? You never gave an answer.

 

[Bashyboy]

I am terribly Sorry, Ray, for not having answered your question. Yes, I am able to use the fact that
a*a = (-a)*(-a)

 

[Ray Vickson]

I am terribly Sorry, Ray, for not having answered your question. Yes, I am able to use the fact that
a*a = (-a)*(-a)

No need to apologize: the question was for your benefit, not mine.