Question

Give a mathematical application of the theorem 2^x > x @ganeshie8 @RadEn @agent0smith @dan815 @shamil98 @chmvijay @nincompoop @AriPotta @Ashleyisakitty @wolfe8 @Mashy @pgpilot326 @jagatuba @RANE @perl @NotTim @tyteen4a03 @divu.mkr @Kristen17 @tanya123 @kittiwitti1 @sourwing @Squirrels @thadyoung @LoverofWords @cebroski @alekos @UH60blackhawk @linh412986 @NaomiBell1997 @Ness9630 @superraymond @TareKKowshiK @Partycool @MeganEdward @Rockiee @David.... @ninskii @RaymondoOnley @alexp0k3r @asib1214 @13xyhuang2 @polaris_s0i @nert
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Answer 1

Well, we could use this theorem to prove that 2^x > x - 1 because x > x - 1, so 2^x > x > x - 1 means that 2^x > x - 1.

Answer 2

Well the answer is CANTOR'S THEOREM! RING A BELL? :))))))))))))))))))))

Answer 3

Can that theorem be proved with induction?

Answer 4

Cantor's theorem was based off the fact that 2^x > x. Do you recall what Cantor's theorem is?

Answer 5

Not at the moment, although I am concerned with how to prove 2^x > x

Answer 6

That's kind of besides the point :P Just graph them on a calculator and you'll see that it's true :P

Answer 7

Proof by induction

Step 1) 2^1 = 2 > 1
Step 2) Assume 2^n > n is true, and assume that n >= 1
2 * 2^n > 2*n
2^(n + 1) > 2n = n + n >= n + 1
2^(n + 1) > n + 1 for all n >= 1

Therefore, 2^x > x

Answer 8

If x = 0, then 2^0 = 1 > 0
If x < 0, then 2^x > 0 > x, so 2^x > x

Answer 9

Would you like to tell me about Cantor's Theorem?

Answer 10

Consider a set X. Cantor's Theorem states that the number of members of a set X will always be less than the members of the power set P(x), which has the convenient formula Members[P(x)] = 2^Members(x)

Answer 11

You don't need to use \(n<2^n\) for that.
If you consider all subsets of \(X\) with cardinality \(1\), that alone is equal to the cardinality of \(X\). We know the power set will also have the empty set making it's cardinality at least greater by \(1\).

So even without knowing the cardinality of the power set in terms of the cardinality of \(X\) and even without knowing that \(n<2^n\), we can trivially see that the power set has more members.

Answer 12

@FutureMathProfessor no mass tagging please and I have no knowledge of this so please tag me in things I can help you with (:

Answer 13

Well @wio, when writing a formal proof you can't just state that. It needs to be rigorous, it may be obvious, but the more obvious something is the more difficult it is to prove in most cases.

Answer 14

@malevolence19 There isn't any lack of rigor in my proof. We know the power set will contain all cardinality \(1\) subsets of \(X\) as well as the empty set. We know there is a one to one mapping between between cardinality \(1\) subsets of \(X\) and elements.

Answer 15

That's better :P But yes, what you said makes sense regardless.