Is a^2 Always Positive for a Real Number a?

[r0bHadz]

Homework Statement

If a is a real number, then a^2 is positive

Homework Equations

The book gives me two propositions:

1- if a,b are positive, so are the product ab and the sum a+b
2- if a is a real number, then either a is positive, a = 0, or -a is positive, and these possibilities are mutually exclusive

The Attempt at a Solution

According to proposition 2 and in my text which reads "we have the positive numbers, represented geometrically on the straight line by those numbers unequal to 0 and lying to the right of 0. if a is a positive number, we write a>0

So a real number can be broken down into positive, negative, and 0.

If a is a positive real number, a^2 is positive by proposition 1, but you have to prove this for all three cases and I don't see how you can possibly prove this for zero.

 

[member 587159]

If your book defines a number $p$ to be positive if $p > 0$, your book is false: $0^2 = 0$ isn't positive.

On the other hand, if it defines a number $p$ to be positive if $p \geq 0$, it is correct.

 

[fresh_42]

It's wrong for zero, so you can't prove it. The correct word is nonnegative.

 

[r0bHadz]

If your book defines a number $p$ to be positive if $p > 0$, your book is false: $0^2 = 0$ isn't positive.

On the other hand, if it defines a number $p$ to be positive if $p \geq 0$, it is correct.

Why would Lang do this? When I look up the answer to the question, he only gives an answer for the case of a positive real number x,x>0

I don't know why he would make the question "if a is a real number" then just disregard the other 2 cases that would constitute real numbers

 

[member 587159]

Why would Lang do this? When I look up the answer to the question, he only gives an answer for the case of a positive real number x,x>0

I don't know why he would make the question "if a is a real number" then just disregard the other 2 cases that would constitute real numbers

Maybe because the case $a = 0$ is trivial (depending on the definition it is either trivial or false) and the case $a < 0$ follows from the case $a > 0$. Indeed, if $a < 0$, then $-a > 0$ and

$$a^2 = (-a)(-a) > 0$$

 

[r0bHadz]

Maybe because the case $a = 0$ is trivial (depending on the definition it is either trivial or false) and the case $a < 0$ follows from the case $a > 0$. Indeed, if $a < 0$, then $-a > 0$ and

$$a^2 = (-a)(-a) > 0$$

Do you suggest reading a different book for basic math? I feel like he should be making things such as a>0 and a=0 being positive clear, but he hasn't. Right now I think giving him another 2 chances would be fair since this is the only major issue I've encountered so far, but ynn

 

[member 587159]

Do you suggest reading a different book for basic math? I feel like he should be making things such as a>0 and a=0 being positive clear, but he hasn't

I didn't read the book you are mentioning, so I am not in a position to say that you should leave the book. I can recommend other books, but you should tell me first:

(1) What is your goal by reading those books?
(2) What is your background?

 

[r0bHadz]

I didn't read the book you are mentioning, so I am not in a position to say that you should leave the book. I can recommend other books, but you should tell me first:

(1) What is your goal by reading those books?
(2) What is your background?

1- I want to be able to complete Spivak's calculus and then move on to study physics relating to electricity and magneticism
2- I am a CS student

 

[fresh_42]

Do you suggest reading a different book for basic math?

If you are talking about Serge Lang, then the answer is no. It is a minor inaccuracy, not really worth to condemn the entire book for.

 

[member 587159]

1- I want to be able to complete Spivak's calculus and then move on to study physics relating to electricity and magneticism
2- I am a CS student

Is Spivak currently too hard? What makes you read Lang's book before Spivak's?

 

[r0bHadz]

Is Spivak currently too hard? What makes you read Lang's book before Spivak's?

The book I am reading has examples that require proof's, and it has examples requiring calculations. I've taken up to calculus 3 but I have never had to prove anything besides some basic delta epsilon proofs.

I was told Spivak's book is proof heavy so I decided to start off with basic math and move up.

 

[r0bHadz]

Prove: if a is negative, then a^(-1) is negative

proof:

If a is negative, -a is positive, and (-a)(a^-1) = (-1)(a)(a^-1) = (-1)(1) = -1

since -a was positive, a^-1 must be negativeIs this proof valid?

 

[fresh_42]

Prove: if a is negative, then a^(-1) is negative

proof:

If a is negative, -a is positive, and (-a)(a^-1) = (-1)(a)(a^-1) = (-1)(1) = -1

since -a was positive, a^-1 must be negativeIs this proof valid?

That depends on what you may use. If you have: "Only the multiplication by a negative number can turn a positive into a negative one." plus "-1 is negative." then yes, Otherwise you will have to prove these two statements first, or find another proof.

You could proceed indirect: Assume $a^{-^1} \geq 0$. Zero isn't possible, so only $a^{-1}>0$ is left to consider. Since $a<0$, we have $-a>0$ by the second rule and $(-a) \cdot a^{-1} = -1 > 0$ by the first rule. This way you only have to prove $-1 < 0$ to obtain a contradiction.

 

[member 587159]

The book I am reading has examples that require proof's, and it has examples requiring calculations. I've taken up to calculus 3 but I have never had to prove anything besides some basic delta epsilon proofs.

I was told Spivak's book is proof heavy so I decided to start off with basic math and move up.

Spivak's book is proof heavy, yes. However, the book can be read as a first and gentle introduction to higher mathematics. Try to read some of it. It might be easier than you think (or harder).

 

[FactChecker]

Good catch. You are right, that the statement you were asked to prove is incorrect for a=0. You should not switch books because of that error in the book. If there is a page of known errors, that one might be included. Proofreading is a hard, thankless, endless job.

 

[SammyS]

Do you suggest reading a different book for basic math? I feel like he should be making things such as a>0 and a=0 being positive clear, but he hasn't. Right now I think giving him another 2 chances would be fair since this is the only major issue I've encountered so far, but ynn

It's quite clear from proposition 2 that: if a = 0, then a is neither positive nor negative.

Homework Equations

The book gives me two propositions:

1- if a,b are positive, so are the product ab and the sum a+b
2- if a is a real number, then either a is positive, a = 0, or -a is positive, and these possibilities are mutually exclusive

 

[r0bHadz]

It's quite clear from proposition 2 that: if a = 0, then a is neither positive nor negative.

Which is why I asked should I consider reading another book, given the fact that he didn't define 0 to be positive then asked me to prove something that would only be possible to prove if 0 was defined as positive

 

[fresh_42]

Which is why I asked should I consider reading another book, given the fact that he didn't define 0 to be positive then asked me to prove something that would only be possible to prove if 0 was defined as positive

It will be hard to find a book without those little inaccuracies, which range more in the region of typos than actual mistakes.

 

[r0bHadz]

It will be hard to find a book without those little inaccuracies, which range more in the region of typos than actual mistakes.

Yeah I don't doubt that. I've decided to continue with the book and hope to be done within a week. I will report back to you guys if I have any questions or concerns, since school is out already..