# triclino

Senior Members

285

1. ## inequality

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

How do we solve the system of the following equations: 1) $x+y+z=0$ 2) $x^2+y^2+z^2 =0$ 3)$x^3+y^3+z^3 =0$
3. ## set of equations

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.
4. ## inequality

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): $\neg(a\Leftrightarrow b)\ \equiv\ (a\wedge \neg b)\vee(\neg a\wedge b)$ To get :$\neg[p\Leftrightarrow(q\Leftrightarrow r)]$ You must have :$[p\wedge\neg(q\Leftrightarrow r)]\vee[\neg p\wedge(q\Leftrightarrow r)]$ Instead of $[p\wedge\neg(q\Leftrightarrow r)]\wedge[\neg p\wedge(q\Leftrightarrow r)]$ That you have in your proof
6. ## set of equations

Solve the following system of equations: 1)$x^2 +2yz =13$ 2)$y^2 +2xz = 13$ 3)$z^2 +2xy =10$
7. ## set of equations

Find the reals x,y,z satisfying the following equations: 1) x+y+z = 0 2) $x^3 + y^3 + z^3 = 3xyz$ 3) xy + yz + xz = $a^2$ when a=0 and when $a\neq 0$
8. ## inequality

I don't get you
9. ## inequality

prove : $a^2+ab+b^2\geq 0$ for all a,b real Nos
10. ## inequality

For what values of a and b does the following inequality hold: $a^2+ab-4a+3>0$

12. ## Peculiar proof

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]
14. ## Peculiar proof

Contradiction is: $q\wedge\neg q$ q here is 1=0 and not q is $1\neq 0$ You thing twice about it $1\neq 0$ is a field axiom (x=0 and x=1) is equivalent to 1=0 ,and not(x=0 and x=1) is equivalent to:$1\neq 0$
15. ## Peculiar proof

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)
16. ## Peculiar proof

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 $|x+y|\leq 2$ ,then : $2<|x+y|\leq 2$ .Which of course results in 2>2
17. ## Peculiar proof

And just how do we do that??
18. ## Peculiar proof

Then how do we treat the 4th case in this new problem??
19. ## Peculiar proof

And if the problem asked ,instead of the statement :$x+y\leq 2$,the statement : $|x+y|\leq 2$ to be proved?? Would you use simplification again??
20. ## Peculiar proof

Your kind of proof is based on what fact(s)?
21. ## Peculiar proof

No, not at all because x=0 and x=1 results in 0=1
22. ## Peculiar proof

In proving : $x+y\leq 2$ given that : $0\leq x\leq 1$ and y=1,the following proof is pursued: Since $0\leq x\leq 1$ ,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 : $x+y\leq 2$. The question now is : how do we examine the 4th case so to end up with $x+y\leq 2$??
23. ## on differentials again

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??
24. ## on differentials again

Sorry again to be exact the function is defined as follows: $f : [0,1]\Longrightarrow R$,where f(x)= 1 ,if x is rational and f(x) =0 ,if x is irrational
25. ## on differentials again

Sorry i meant : f(x)= 1 if x is rational and, f(x) =0 if x is irrational . I corrected my post
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