triclino

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.

How about, b . For what values of b does the inequality hold??

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]

Where is the 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]

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]

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)

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

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]??

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