Search results

1. Prove Divisibility

I like LaTEX. Why do you hate it?
2. Epsilon- Delta Proof

Homework Statement Prove that ## lim_{x\implies 1} \frac{2}{x-3} = -1 ## Use delta-epsilon. The Attempt at a Solution Proof strategy: ## | { \frac{ 2}{x-3} +1 } | < \epsilon ## ## \frac{x-1}{x-3} < \epsilon ## , since delta have to be a function of epsilon alone and not include x. I...

Thanks.
4. Reflexivity Implies Symmetry?

Wait, so my R is an equivalence relation then? Supposedly, it partitions the set into disjoint classes. I guess that my classes would be [1], [2], [3]? HallsofIvy, thank you for the clarification. I should have stated that in this case, it means the same.
5. Reflexivity Implies Symmetry?

Nevermind. I just read somewhere that reflexive statements don't count towards symmetry. Apparently, it involves something like a diagonal class; I guess they pair this combinations in a matrix like form. Anyway. Thanks.
6. Reflexivity Implies Symmetry?

Homework Statement Is this relation, R, on ## S= \{ 1, 2, 3 \} \\ R = \{ (1,1), (2,2) , (3,3) \}## Symmetric? It is obvious that it is reflexive.
7. Show F is Injective & Cardinality of Domain

Now, I am wondering... do they have to be equal? a =2 and b =3. If the only restriction is that a,b ## \in \mathbb{N}## What would be an argument against what I just wrote?
8. Prove Divisibility

Sorry, I got confused. Let's see: ## a\not | c \wedge a\not | b \implies a \not |bc ## ## pr \not = pqx \wedge qr \not = pqy \implies pq(r^2) \not = pqz## ## T \wedge T \implies F \equiv F## So, the statement is false.
9. Show F is Injective & Cardinality of Domain

To show that ## g## is injective: ## g(a) = g(b ) \\ (a,1 ) = (b ,1 )## Then, ##a## must be equal to ## b## and ## g ## is injective. Thank you for the unique factorization explanation.
10. Prove Divisibility

You are right, it is sufficient. Given your conditions, the converse technically would be true, since ##(F \wedge F \implies F) \equiv T## Am I right?
11. Co-Primes Proof

Thank you very much, Curious. BTW, that cat in your pic looks happy :>
12. Prove Divisibility

Homework Statement a.) Prove: If an integer ##a## does not divide ##bc##, then ##a## does not divide ##b## and ##a## does not divide ##c##. b.) State and either prove or disprove the converse of the above statement. The Attempt at a Solution a.) Proof by contrapositive ## a|c \vee a|b...
13. Co-Primes Proof

So, d | n d =3 does 3 |n ? Well, the stipulations say that ##n## is not divisible by 3. ## n\not | 3 ## Sorry, but again I am lost. ## (n = 3k )\not = (3 \not = np), k,p \in \mathbb{Z} ##
14. Can I use induction?

I see what you are doing. About what I wrote prior to this... I was talking about this supposedly property of absolute value ## |a-b| < c ##, then ## -c< a-b <c ##
15. Can I use induction?

Wouldn't I need this first: ## -\sqrt{ a^2 + b^2} < (a-b) < \sqrt{ a^2 +b^2 }\\ ## Then, ## a^2 +b^2 < ( a-b )^2 = (a^2 - 2ab + b^2 ) < (\sqrt{a^2 +b^2})^2 = a^2 +b^2## Now, I don't understand
16. Can I use induction?

uh, are you serious? ## (|a-b|)^2 = (a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2 = a^2 +b^2## Then the LHS will always be lesser for any ## \mathbb{R^+}##
17. Can I use induction?

Homework Statement Prove for any ##a,b \in \mathbb{R^+} : |a-b| \leq \sqrt{a^2 +b^2} ## The Attempt at a Solution How should I start? Can I use induction? Should I use contrapositive?
18. Co-Primes Proof

So, ## d=3 \\ n =3k \\ n+3 =3p \\ gcd(n, n+3 ) =1 \\ nh + (n+3)x =1, h,x \in \mathbb{Z} \\ 3(kh+px) =1, kh+px \in \mathbb{Z} ## Then, 3|1; however, 1 =3y is impossible, since y must be integer. So, necessarily d =1 only. I still don't see the connection. Are we assuming that ## d...
19. Co-Primes Proof

Since 3 is prime d can be 1 or 3.
20. Co-Primes Proof

## d| n+3 \\ d|n \\ n =dk \wedge n +3 = dp, p,k \in \mathbb{Z} \\ 3=d(p-k) ## Since p-k is an integer, d|3 ? I don't get where is the connection of ##d## with ##n## not divisible by 3.
21. Co-Primes Proof

Thanks. You mean state Euclid's lemma? Let ## a,b,c \in \mathbb{Z} \wedge a \not = 0 ## . If ## a|bc \wedge gcd(a,b) =1 , ##then ## a|c ## ## n | (n+3)c , c \in \mathbb{Z} \\ n| c##?
22. Co-Primes Proof

Homework Statement Let ##n \in \mathbb{Z} , n \not | 3##. Prove that ##gcd( n , n +3 ) =1 ## The Attempt at a Solution If n is not divisible by 3, then n = 3k+1 or n =3k+2 , ## k \in \mathbb{Z} ## What is a feasible approach? Can I do this? For first case, ## gcd(3k+1, 3k+4 ) = 1 \\...
23. Show F is Injective & Cardinality of Domain

You are right. Just to make this clear to myself, I will define g as strictly injective and very non-surjective. g(n) = (n, 1), where ( n,1) ## \in S ##
24. Show F is Injective & Cardinality of Domain

I have to show an injective relation from N into S. Can I say that g(x,y) = (x,y), where x and y ## \in \mathbb{N} ## Therefore, by Cantor, Schroder, Bernstein's theorem ## \mathbb{N} \wedge S ## have the same cardinality.
25. Prove Existence of Real Number

Thanks, Kurtz! ## (x^2 - 1)^2 = -1 ##, since the square of any number is always non-negative, it is impossible that x is real.
26. Show F is Injective & Cardinality of Domain

I could state in another way, but I still have no argument as to why they are equal. I was thinking that I could say ## 2^{a-c} = 3^{ d-b} ## and somehow since there is not a power other than 0 that satisfies then a =c and d =b. But, I am not sure what is this "somehow" or if it is even...
27. Prove Existence of Real Number

## \not\exists x | x \in \mathbb{R}, ## since ## x = \sqrt{x^2} = \sqrt{ -1 +- \sqrt{-1} }, x \in \mathbb{C}##
28. Show F is Injective & Cardinality of Domain

Homework Statement Let ## S = \{ (m,n) : m,n \in \mathbb{N} \} \\ ## a.) Show function ## f: S -> \mathbb{N} ## defined by ## f(m,n) = 2^m 3^n ## is injective b.) Use part a.) to show cardinality of S. The Attempt at a Solution a.) ## f(a,b) = f(c, d ) ; a,b,c,d \in \mathbb{N} \\\\ 2^a...
29. Prove Existence of Real Number

Can I do this to the quadratic equation? : ## x^4 - 2x^2 +2 = 0 \\ (x^2)^2 - 2(x^2) +2 =0 \\ ## And, I solve for ## x^2 = -1 +- \sqrt{-1}##
30. Prove Existence of Real Number

I said 1 or -1 because I was trying the base case after 0, but yeah let me try the quadratic formula.