# Homework Help: Translating Statements

1. Jul 5, 2013

### Bashyboy

1. The problem statement, all variables and given/known data
Express each of these mathematical statements using predicates, quantifiers, logical connectives, and mathematical operators.

c) Every positive real number has exactly two square roots.

d) A negative real number does not have a square root that is a real number.

2. Relevant equations

3. The attempt at a solution

For part c), the answer in the textbook is, $\forall x \exists a \exists b(a ≠ b \wedge \exists c(c² = x ↔ (c = a \vee c = b)))$ my answer is, $\forall x \exists y \exists z [((x > 0) \wedge (a \ne b)) \implies ((\sqrt{x} = y) \wedge (\sqrt{x} = z))]$

For part d), the answer in the textbook is, $\forall x ((x < 0) \implies \neg \exists y (x = y^2))$; my answer is, $\forall x[(x < 0) \implies (\sqrt{x} \notin \mathbb{R})$

I was wondering, are my answers as valid as the ones provided in the textbook?

Last edited: Jul 5, 2013
2. Jul 5, 2013

### Staff: Mentor

Your answer to (c) introduces y and z, but uses a and b afterwards?
You cannot use the square root symbol here - it is a function with only one value (the positive root). It cannot be a AND b at the same time, as you want a!=b.
You probably don't want the effect of "=>" in your formula. It is satisfied even if both sides are wrong (with a=b, for example), so it does not make any statement about the existence of roots at all.

For (d): You cannot write $\sqrt{x}$ if you don't know if that is defined at all. With complex numbers, that is possible, but even then you need a definition of the square root function first.

3. Jul 5, 2013

### Bashyboy

Okay, well I understand the textbook's answer for part d); however, I am having a little more difficulty with part c). My question is, where does it answer specify that x is positive? And to be clear, a and b are the two roots that are mentioned in the original statement?

4. Jul 5, 2013

### Staff: Mentor

That is the magic of "<=>". For negative x, there is no such c, both sides are false and the equality holds. I would write $\forall x \in \mathbb{R}^+$, however.

$\in \mathbb{R}$ for all symbols is missing.

5. Jul 5, 2013

### verty

I think it has been assumed that all variables range over the real numbers. For this reason, (d) is incorrect as it assumes the square root of x is real for any candidate x.

I didn't notice that the book's answer for (c) is wrong. I don't particularly like the book's answer anyhow. I think I would use this logic: x has two distinct roots and any third root is non-distinct.

Bashyboy: see if you can translate my statement.

6. Jul 5, 2013

### haruspex

I can't make much sense of that. Are you sure you've copied this correctly?

7. Jul 8, 2013

### Bashyboy

Haruspex, that is actually copied-and-pasted from my electronic textbook, so there can't possible exist any error in transcribing.

8. Jul 8, 2013

### haruspex

Well, it's wrong. It should be $\forall x \exists a \exists b(a ≠ b \wedge \forall c(c² = x ↔ (c = a \vee c = b)))$. Note the $\forall c$.

9. Jul 8, 2013

### Bashyboy

Ah, yes, that is in fact more comprehensible.