an equality

[chrislav]

prove the following:

$$a\geq 0\wedge b\geq 0 \implies a^2=b^2\implies a=b$$
proof:
(A) $$a^2=b^2\implies (a+b)=0\vee (a-b)=0$$
$$a\geq 0\wedge b\geq 0$$ =>[ (a>o or a=0)and (b>0 or b=0)] => [(a>o and b>o)or (a>0 and b=0) or (a=0 and b>0) or(a=0 and b=0)]

so we have:

1)(a>o and b>o) => a+b>0 => ¬(a+b=0) and using (A) we have a=b

2)(a>0 and b=0) => a+b=0 =a+0=0 => a=0=b => a=b

3) (a=0 and b>0) => a+b=0=0+b=0 => b=0=a => a=b

4)(a=0 and b=0) => a=0=b => a=b

Hence $$a^2=b^2\implies a=b$$
Is that proof correct?

 

[fresh_42]

prove the following:

$$a\geq 0\wedge b\geq 0 \implies a^2=b^2\implies a=b$$
proof:
(A) $$a^2=b^2\implies (a+b)=0\vee (a-b)=0$$
$$a\geq 0\wedge b\geq 0$$ =>[ (a>o or a=0)and (b>0 or b=0)] => [(a>o and b>o)or (a>0 and b=0) or (a=0 and b>0) or(a=0 and b=0)]

so we have:

1)(a>o and b>o) => a+b>0 => ¬(a+b=0) and using (A) we have a=b

2)(a>0 and b=0) => a+b=0 =a+0=0 => a=0=b => a=b

3) (a=0 and b>0) => a+b=0=0+b=0 => b=0=a => a=b

4)(a=0 and b=0) => a=0=b => a=b

Hence $$a^2=b^2\implies a=b$$
Is that proof correct?

Yes.

However, I think the many cases can be shortened a bit. You also should use an additional pair of parentheses:
$$a\geq 0\wedge b\geq 0 \implies \left(a^2=b^2\implies a=b\right).$$The idea to write $a^2-b^2=(a-b)\cdot (a+b)=0$ is perfect. But now, we can use the property that $x\cdot y=0 \implies x=0 \vee y=0$ for any field of numbers that contains the rational numbers, e.g. the field of real numbers. This property holds whenever $\underbrace{1+1+\ldots+1}_{n\text{ times}}\neq 0$ for all positive integers $n.$ It immediately results in
$$0=a^2-b^2=(a-b)\cdot (a+b) \implies (a-b=0) \vee (a+b=0).$$If $a+b=0$ then $a=-b.$ Since both numbers have the same sign, this can only hold if $a=b=0.$ Otherwise, $a-b=0 .$ We thus have $a=b$ in both cases.

The property $\underbrace{1+1+\ldots+1}_{n\text{ times}}\neq 0$ for all positive integers $n$ does not always hold. For the electronic device you currently use while reading this, or the light switch in your room holds $1+1=0.$

 

[chrislav]

Yes.

However, I think the many cases can be shortened a bit. You also should use an additional pair of parentheses:
$$a\geq 0\wedge b\geq 0 \implies \left(a^2=b^2\implies a=b\right).$$The idea to write $a^2-b^2=(a-b)\cdot (a+b)=0$ is perfect. But now, we can use the property that $x\cdot y=0 \implies x=0 \vee y=0$ for any field of numbers that contains the rational numbers, e.g. the field of real numbers. This property holds whenever $\underbrace{1+1+\ldots+1}_{n\text{ times}}\neq 0$ for all positive integers $n.$ It immediately results in
$$0=a^2-b^2=(a-b)\cdot (a+b) \implies (a-b=0) \vee (a+b=0).$$If $a+b=0$ then $a=-b.$ Since both numbers have the same sign, this can only hold if $a=b=0.$ Otherwise, $a-b=0 .$ We thus have $a=b$ in both cases.

The property $\underbrace{1+1+\ldots+1}_{n\text{ times}}\neq 0$ for all positive integers $n$ does not always hold. For the electronic device you currently use while reading this, or the light switch in your room holds $1+1=0.$

how do you know that my proof i correct?
how do you prove:

If $a+b=0$ then $a=-b.$ Since both numbers have the same sign, this can only hold if $a=b=0.$

how do you prove that?

 

[chrislav]

i may add that the solution of the op must be witout using sqrs roots

 

[fresh_42]

how do you know that my proof i correct?

Well, you dealt with all possible cases. If we are picky, then cases 2) and 3) are at least not correctly phrased. If one number is positive and the other one is zero, you cannot conclude $a=b.$ You should have written that this case is impossible because it would follow that $a^2-b^2>0$ contradicting the given condition $a^2-b^2=0 .$ This uses the fact that the real (or rational) numbers are totally ordered and therefore the ordering obeys the law of trichotomy.

how do you prove:

how do you prove that?

If $a+b=0$ then $a=-b$ follows from the fact that the real (or rational) numbers, or any field build an additive group. It follows from the group axioms that the additive inverse is unique. Since $a+b=0$ and $a+(-a)=0$ uniqueness implies $b=-a.$

The second step follows again from the law of trichotomy. If $b\geq 0$ then $-b\leq 0.$ Simply add $-b$ on both sides. Now, if $a=-b\geq 0$ and $a=-b\leq 0$ then $a=-b=0$ and $b=0.$

Square roots are irrelevant here. Only the group axioms and the total ordering is used. The statement would be false for complex numbers as they are not totally ordered. However, you did neither specify where $a,b$ are taken from, nor which properties of this domain were allowed to use, so I had to guess. My guess was real numbers together with the usual properties: field, characteristic zero, totally ordered.

 

[chrislav]

Well, you dealt with all possible cases. If we are picky, then cases 2) and 3) are at least not correctly phrased. If one number is positive and the other one is zero, you cannot conclude $a=b.$ You should have written that this case is impossible because it would follow that $a^2-b^2>0$ contradicting the given condition $a^2-b^2=0 .$ This uses the fact that the real (or rational) numbers are totally ordered and therefore the ordering obeys the law of trichotomy.

if (a>0 and b=0) =>(b=0). Then a+b=0 => a+0=0=>a=0=>a=0=b=>a=b..Because if you put : a>o as p and b=0 as q then p&q logicaly implies q which is b=0 .This law of logic is called conjunction elimination in propositional calculus

If $a+b=0$ then $a=-b$ follows from the fact that the real (or rational) numbers, or any field build an additive group. It follows from the group axioms that the additive inverse is unique. Since $a+b=0$ and $a+(-a)=0$ uniqueness implies $b=-a.$

The second step follows again from the law of trichotomy. If $b\geq 0$ then $-b\leq 0.$ Simply add $-b$ on both sides. Now, if $a=-b\geq 0$ and $a=-b\leq 0$ then $a=-b=0$ and $b=0.$

Square roots are irrelevant here. Only the group axioms and the total ordering is used. The statement would be false for complex numbers as they are not totally ordered. However, you did neither specify where $a,b$ are taken from, nor which properties of this domain were allowed to use, so I had to guess. My guess was real numbers together with the usual properties: field, characteristic zero, totally ordered.

How do you prove that:

if -b is gteater or equal to zero and also -b is less or equal to zero ,then -b=0

 

[fresh_42]

if (a>0 and b=0) =>(b=0). Then a+b=0 => a+0=0=>a=0=>a=0=b=>a=b..Because if you put : a>o as p and b=0 as q then p&q logicaly implies q which is b=0 .This law of logic is called conjunction elimination in propositional calculus

Whatever you call it. But $a=0$ is a contradiction to $a>0.$ You cannot add $a=b$ since you already have a false statement in your chain of conclusions. After the contradiction $a> 0 \wedge a=0$ you have to stop since from that point on, literally every statement is true. From FALSE follows TRUE AND FALSE, hence every statement after FALSE is logically void. In any case, $a=b$ cannot be concluded in cases 2) and 3). You already showed that no such $a$ exists, hence a comparison with $b$ is a void statement, which by the way is always true. However, $a\neq b$ would also be true.

The contradiction itself is the violation of the law of trichotomy.

How do you prove that:

if -b is gteater or equal to zero and also -b is less or equal to zero ,then -b=0

By the law of trichotomy that follows from the total ordering. It says that any real number is either negative, or positive, or zero. Hence
$$\begin{array}{lll} \left[ (-b) \geq 0 \right] \text{ AND } \left[ (-b) \leq 0 \right] &\Longleftrightarrow \left[ \left[ (-b) > 0 \right] \text{ OR } \left[ (-b)=0 \right] \right] \text{ AND } \left[ \left[ (-b) < 0 \right] \text{ OR } \left[ (-b)=0 \right] \right] \\ &\Longleftrightarrow \left[ (-b)=0 \right] \text{ AND } \left[ (-b)=0 \right] \\ &\Longleftrightarrow (-b)=0 \end{array}$$

 

[chrislav]

which laws of logic are used in the ops proof

Whatever you call it. But $a=0$ is a contradiction to $a>0.$ You cannot add $a=b$ since you already have a false statement in your chain of conclusions. After the contradiction $a> 0 \wedge a=0$ you have to stop since from that point on, literally every statement is true. From FALSE follows TRUE AND FALSE, hence every statement after FALSE is logically void. In any case, $a=b$ cannot be concluded in cases 2) and 3). You already showed that no such $a$ exists, hence a comparison with $b$ is a void statement, which by the way is always true. However, $a\neq b$ would also be true.

The contradiction itself is the violation of the law of trichotomy.



By the law of trichotomy that follows from the total ordering. It says that any real number is either negative, or positive, or zero. Hence
$$\begin{array}{lll} \left[ (-b) \geq 0 \right] \text{ AND } \left[ (-b) \leq 0 \right] &\Longleftrightarrow \left[ \left[ (-b) > 0 \right] \text{ OR } \left[ (-b)=0 \right] \right] \text{ AND } \left[ \left[ (-b) < 0 \right] \text{ OR } \left[ (-b)=0 \right] \right] \\ &\Longleftrightarrow \left[ (-b)=0 \right] \text{ AND } \left[ (-b)=0 \right] \\ &\Longleftrightarrow (-b)=0 \end{array}$$

you are confusing (a>0 and b=o) with a>0 Also nowhere in the proof you will a>0 on its own
You wil find a>0 or equal to o OR (a>o and b>0), (a>0 and b=0) or (a=0 and b>0) or(a=0 and b=0)
NOWHERE YOU WILL FIND( A>O AND A=0) BUT YOU WILL FIND (A>0 OR A=0)
WILL CHECK YOUR PROOF TOMORROW
(-B>0 0R B=O) AND (-B<0 OR -B=0)=>-B=0
But keep in mind that we have to use the distributive laws of propositional calculus