triclino

You should find that the determinant is negative if b is within a certain range of values. I make it [hide]4−2√3< b < 4+2√3[/hide] In this case, the inequality is satisfied for all real values of a.

How about a=0 .Is not the inequality satisfied for all values of b??

How do we solve the system of the following equations:

1) [math] x+y+z=0[/math]

2) [math] x^2+y^2+z^2 =0[/math]

3)[math] x^3+y^3+z^3 =0[/math]

It is understandable that the problems that I bring to the attention of the forum are problems that i cannot solve .

Hence the expression :" solve this"

There also problems that i have an idea of their proof but i am not quite sure .

Definitely none of them are homework since i am not a candidate of any educational institute neither am i planning to become one.

If the determinant is negative (treating the LHS as a quadratic in a), then the inequality holds for all real values of a.

How about, b .

For what values of b does the inequality hold??

Use the following rules:

 

[math]a\Leftrightarrow b\ \equiv\ (a\wedge b)\vee(\neg a\wedge \neg b)[/math]

 

[math]\neg(a\Leftrightarrow b)\ \equiv\ (a\wedge \neg b)\vee(\neg a\wedge b)[/math]

 

Thus: [math]\neg[(p\Leftrightarrow q)\Rightarrow r][/math]

 

[math]\implies\ (p\Leftrightarrow q)\wedge\neg r[/math]

 

[math]\implies\ [(p\wedge q)\vee(\neg p\wedge\neg q)]\wedge\neg r[/math]

 

[math]\implies\ [(p\wedge q)\wedge\neg r]\vee[(\neg p\wedge\neg q)\wedge\neg r][/math]

 

[math]\implies\ [p\wedge(q\wedge\neg r)]\vee[\neg p\wedge(\neg q\wedge\neg r)][/math]

 

[math]\implies\ \{[p\wedge(q\wedge\neg r)]\vee[\neg p\wedge(\neg q\wedge\neg r)]\}\vee\{[p\wedge(\neg q\wedge r)]\vee[\neg p\wedge(q\wedge r)]\}[/math]

 

[math]\implies\ \{[p\wedge(q\wedge\neg r)]\vee[p\wedge(\neg q\wedge r)]\}\vee\{[\neg p\wedge(q\wedge r)]\vee[\neg p\wedge(\neg q\wedge\neg r)]\}[/math]

 

[math]\implies\ \{p\wedge[(q\wedge\neg r)\vee(\neg q\wedge r)]\}\wedge\{\neg p\wedge[(q\wedge r)\vee(\neg q\wedge\neg r)]\}[/math]

 

[math]\implies\ [p\wedge\neg(q\Leftrightarrow r)]\wedge[\neg p\wedge(q\Leftrightarrow r)][/math]

 

[math]\implies\ \neg[p\Leftrightarrow(q\Leftrightarrow r)][/math]

So now let us examine your proof:

According to your rule (which you must show how you get):

[math]\neg(a\Leftrightarrow b)\ \equiv\ (a\wedge \neg b)\vee(\neg a\wedge b)[/math]

To get :[math] \neg[p\Leftrightarrow(q\Leftrightarrow r)][/math]

You must have :[math] [p\wedge\neg(q\Leftrightarrow r)]\vee[\neg p\wedge(q\Leftrightarrow r)][/math]

Instead of [math] [p\wedge\neg(q\Leftrightarrow r)]\wedge[\neg p\wedge(q\Leftrightarrow r)][/math]

That you have in your proof

Solve the following system of equations:

1)[math]x^2 +2yz =13[/math]

2)[math]y^2 +2xz = 13[/math]

3)[math]z^2 +2xy =10[/math]

Find the reals x,y,z satisfying the following equations:

1) x+y+z = 0

2) [math] x^3 + y^3 + z^3 = 3xyz[/math]

3) xy + yz + xz = [math]a^2[/math] when a=0 and when [math]a\neq 0[/math]

I don't get you

prove : [math] a^2+ab+b^2\geq 0[/math] for all a,b real Nos

For what values of a and b does the following inequality hold:

[math]a^2+ab-4a+3>0[/math]

Use the following rules:

 

[math]a\Leftrightarrow b\ \equiv\ (a\wedge b)\vee(\neg a\wedge \neg b)[/math]

 

[math]\neg(a\Leftrightarrow b)\ \equiv\ (a\wedge \neg b)\vee(\neg a\wedge b)[/math]

 

Thus: [math]\neg[(p\Leftrightarrow q)\Rightarrow r][/math]

 

[math]\implies\ (p\Leftrightarrow q)\wedge\neg r[/math]

 

[math]\implies\ [(p\wedge q)\vee(\neg p\wedge\neg q)]\wedge\neg r[/math]

 

[math]\implies\ [(p\wedge q)\wedge\neg r]\vee[(\neg p\wedge\neg q)\wedge\neg r][/math]

 

[math]\implies\ [p\wedge(q\wedge\neg r)]\vee[\neg p\wedge(\neg q\wedge\neg r)][/math]

 

[math]\implies\ \{[p\wedge(q\wedge\neg r)]\vee[\neg p\wedge(\neg q\wedge\neg r)]\}\vee\{[p\wedge(\neg q\wedge r)]\vee[\neg p\wedge(q\wedge r)]\}[/math]

 

[math]\implies\ \{[p\wedge(q\wedge\neg r)]\vee[p\wedge(\neg q\wedge r)]\}\vee\{[\neg p\wedge(q\wedge r)]\vee[\neg p\wedge(\neg q\wedge\neg r)]\}[/math]

 

[math]\implies\ \{p\wedge[(q\wedge\neg r)\vee(\neg q\wedge r)]\}\wedge\{\neg p\wedge[(q\wedge r)\vee(\neg q\wedge\neg r)]\}[/math]

 

[math]\implies\ [p\wedge\neg(q\Leftrightarrow r)]\wedge[\neg p\wedge(q\Leftrightarrow r)][/math]

 

[math]\implies\ \neg[p\Leftrightarrow(q\Leftrightarrow r)][/math]

Where is the contradiction

A contradiction is any statement that is unsatisfiable.

 

x=0 and x=1 is unsatisfiable. It is a contradiction.

You are entangling False with contradiction .

x=0 and x=1 is simply a false statement.

This is basic stuff in Symbolic Logic, you better learn about it .

Prove using contradiction the following:

[p<=>(q <=>r)] =>[(p<=>q)=> r]

The contradiction here is x=0 and x=1. Think about it.

Contradiction is: [math]q\wedge\neg q[/math]

q here is 1=0 and not q is [math]1\neq 0[/math]

You thing twice about it

[math]1\neq 0[/math] is a field axiom

(x=0 and x=1) is equivalent to 1=0 ,and not(x=0 and x=1) is equivalent to:[math]1\neq 0[/math]

No you can't, triclino. x=0 and x=1 implies absolutely nothing. .

On the contrary D.H contradiction can imply everything.

Thru disjunction syllogism as Tree pointed out you can imply everything.

Learn disjunction syllogism.

The contradiction here is: (x=0 and x=1) and not( x=0 and x=1)

Roughly:

 

Taking not(x=0 and x=1) as given,

 

( (x>0 and x=1) or ( x=0 and x=1) ) and not(x=0 and x=1)

=>

(x>0 and x=1)

 

by disjunctive syllogism.

On the other hand i could use disjunctive syllogism and end up with : 2>2

Since x=0 and x=1 => 0=1 => (0=1) or |x+y|>2 and not (0=1) => |x+y|>2.

But since [math]|x+y|\leq 2[/math] ,then : [math]2<|x+y|\leq 2[/math] .Which of course results in 2>2

. More formally, demonstrate that the first three cases are the only ones that can actually exist.

And just how do we do that??

Then how do we treat the 4th case in this new problem??

And if the problem asked ,instead of the statement :[math]x+y\leq 2[/math],the statement : [math]|x+y|\leq 2[/math] to be proved??

Would you use simplification again??

Your kind of proof is based on what fact(s)?

No, not at all because x=0 and x=1 results in 0=1

In proving : [math] x+y\leq 2[/math] given that : [math]0\leq x\leq 1[/math] and y=1,the following proof is pursued:

Since [math]0\leq x\leq 1[/math] ,then (x>0 or x=0)and (x<1 or x=1) ,which according to logic is equivalent to:

(x>0 and x<1) or (x>0 and x=1) or (x=0 and x<1) or ( x=0 and x=1).

And in examining the three cases ( except the 4th one : x=0 and x=1) we end up with :

[math] x+y\leq 2[/math].

The question now is : how do we examine the 4th case so to end up with [math] x+y\leq 2[/math]??

Oops. I misread your post. I thought you said f(x) = 1 if x is irrational, 0 if x is rational. That function integrates to F(x)=x+c. That is the standard gotcha problem used to introduce Lebesgue integration.

 

You switched the values so that f(x)=1 if x is rational, 0 if x is irrational. This function is zero almost everywhere. It integrates to a constant.

Which is then the function F(x) whose derivative is equal to our function f(x),

so that the differential f(x)dx is exact??

Sorry again to be exact the function is defined as follows:

[math] f : [0,1]\Longrightarrow R[/math],where

f(x)= 1 ,if x is rational and

f(x) =0 ,if x is irrational

Sorry i meant :

f(x)= 1 if x is rational and,

f(x) =0 if x is irrational .

I corrected my post