Inequality Mathematical Induction Proof: 2^n greater than n^2

[@leecoates]

I love this channel. Im an aspiring mathematician and frequently encounter overwhelming self-doubt about my ability. But when you explain something and reassure the audience that you struggled also, it is uplifting to know that it is not just me struggling with seemingly easy concepts. Seriously, thank you so much for this.

[@Dottedshine]

I just had an assignment due today, containing this exact problem. This is a very clear way of explaining it!

[@TheMathSorcerer]

Oh wow what a coincidence!!

[@ChandanKSwain]

@Kathleen McKenzie yes there mentioned that n>4, but he put k=4 , in the middle equation.....

[@bigmansanister8716]

@@ChandanKSwain yea, that confused me as well

[@leonbehrndt2611]

@@TheMathSorcerer Same haha

[@pranjalsrivastava3343]

he discovered gravity xD

[@aminakhan1195]

KZbin SHOULD OPEN A SCHOOL FOR ALL THE KZbin TEACHERS THAT TEACH BETTER THAN SCHOOL TEACHERS. PERIOD.

[@mr.knowitall5019]

@SteveEarl Watt?

[@beri4138]

@Eyosias Tewodros Are you a robot?

[@schizoframia4874]

My ears hurt 🩸

[@syremusic_]

This is an extremely good video because you stumbled (or pretended to :) ) a couple times and talked us through how you figured it out. That’s super helpful. Thank you.

[@TheMathSorcerer]

😄

[@TheMathSorcerer]

Thx😄

[@amacommey192]

You amazed me. I just came from Eddie woo and others for this question now YOU! It’s like you understand the most basic intuition needed to solve it and you did it in so little steps. your solution is gold man. even for the factorial question. THANK YOU SO MUCH

[@TheMathSorcerer]

Aww thank you!!

[@raymondphiri8587]

So true!!!! You are good🙏🏽🙏🏽

[@Hello_am_Mr_Jello]

Same, just came from Eddie

[@Qubit313]

same

[@mathnerdatsdsu6149]

For clarification, I know I am very late to responding to this video, however, when you use k=4 you must be sure that the inductive hypothesis hold for that value of k. If you plug 4 into inductive hyp it actually fails to be true. You must use a value for k that you know the inductive hyp holds true for. In this case it would need to be k=5.

[@xreiiyoox]

yes exactly, that's the part i was confused at to why he put k= 4 when k is bigger than 4, your comment clarified me thanks

[@ayeyukhine466]

@@xreiiyoox I think he puts 4 because of < before k^2.

[@jimpim6454]

What are you talking about its an inequality he didnt 'plug in k=4' he replaced it! I e he threw it in the bin and replaced it with something we know for a fact is smaller than k . Since k is bigger than 4 replacing k with 4 forces an inequality it is him reshaping it so it ends up looking like the conclusion.

[@TomRussle]

@@jimpim6454 yesss such a great descriptive explanation thank you!

[@jimpim6454]

@@TomRussle no problem 😁

[@luuu_na35]

7:31 "Boom" the moment of enlightenment.

[@eguineldo]

I've been struggling a great deal in my proofs class and was self-conscious about my ability to think critically because of it. After watching this, not only do I understand the concept, I feel that I have a greater understanding of how a proof proves its claim. Thank you so much for this video, it has helped immensely!!!

[@渋谷区玲子]

My prof had 1hr and 30 mins to explain this topic and you nailed it within 9 mins. I understood your explanation better than my prof.

[@TheMathSorcerer]

Thx, this is a hard topic to explain! I remember learning this myself and just not getting it. I ended up giving up and only understood it a year later when I looked at it again.

[@Amantheparadise]

@@TheMathSorcerer looking again,is also a mathematical step ,it works

[@aminakhan1195]

Everything's so clear now that I wanna cry oml! THANK YOU!

[@TheMathSorcerer]

You're welcome!!

[@doctorscalling9479]

My teacher tried to prove instead that the difference between the inequalities is bigger than zero, I myself find that much more confusing so when I saw this, I was able to solve any problem of inequalities, thanks alot you are going to save my grades.

[@brandonnorris174]

Thank you for explaining your previous struggles with this kind of proof when you were learning. It really makes the lesson a lot more clear.

[@Andwoo101]

This was amazing. Thank you so much. This is the 8th place I visited trying to find an intuitive explanation.

[@TheMathSorcerer]

Excellent, glad I helped😃

[@jonmartin3026]

Thank you. Hearing you talk through your thought process really helps me understand how to do this myself.

[@schizoframia4874]

I applied an inductive hypothesis for the original induction hypothesis and it seemed to work better

[@Chrisymcmb]

Thank you so much! You really inspire to continue on with school through this math stuff. Sometimes I feel very unmotivated with math because I'll try and I'll try, and when I get it, it's awesome. Plus it's something I genuinely enjoy, so it sucks sometimes when something is just not clicking. Anyhow, I've been watching some of your videos apart from the instructional math ones and they're definitely inspirational, thanks!

[@nkeuphonium]

I appreciate the intuitive approach you take - so much of PMI instruction involves chaotic jumps in reasoning that are hard for listeners to follow and seemingly impossible to intuit ("how did you know to do that?"), so your decision to work with a problem you didn't already know is a great help. :) I was able to get this one a different way, but I had to use a pretty ugly derivative in the middle; your method is much more elegant.

[@757Media]

That was so cool. I am barely starting my classes for my degree and I understood nothing, but it was very cool seeing you work out the problem. Some day I’ll get it.

[@okohsamuel314]

vashTX ... U said "some day I'll get it" ... meaning, u still haven't gotten it.

[@atuly7689]

never saw a more enthusiastic teacher on youtube 👍

[@Art-fn7ns]

Fantastic! Around 5:00 you managed to easily explain what our professor has been failing to...

[@TheMathSorcerer]

👍

[@akindantagonist8708]

Love your channel. So laid back and cool. Helping me so much with my math major. Tysm!

[@TheMathSorcerer]

You are so welcome!

[@raghad5270]

genius! I don't know how to thank you, I was in a trouble and this video saved me, a lot of thanks again..

[@TheMathSorcerer]

You are welcome😃

[@manojtomer0709]

As the problem says n > 4, should we not use 5 instead of 4 in the inductive step? At 6:42

[@hypnogri5457]

How can you replace the k by 4 if it has to be >4?

[@DodiHD]

he messed up there but k^2 + 2k + 10 is still > k^2 + 2k + 1.

[@marangelitorres4515]

@@DodiHD I don't think he messed up. He is not saying k=4, the inequality says > k^2+kk, so whatever is on the left side is greater than this. So using 4, we are saying it will be greater than the value obtained when substituting 4.

[@CallBlofD]

How you know for sure that it will be greater from the value obtained after substituting with 4?

[@isittrueisitnot3303]

I think it goes n>=4 because we had the exact same task like this it was only n>=5 so it's probably a mistake he didn't notice but still correct..

[@nyashadzashegava9568]

'CAUSE K》4.

[@bubbaganoosh]

This was extremely helpful after weeks of struggling. Thank you very much. :D

[@TheMathSorcerer]

Excellent!

[@sanaalshaar5406]

Thank you. I was stuck on the '2^1 x 2^k' for a really, really long time. Induction is tough, and I am really overwhelmed but this video has helped me feel better.

[@rdguezc]

great video!! thanks for sharing your knowledge. I have a question related to the substitution done in the minute 5:00 of the video. You said that "..you allow to do that (the substitution of 2^k by k^2) in math" and change the '=' symbol by '>'. I really want to understand how this substitution is possible and I want to know if you could provide us with any reference or material in which we could go deeper into this subject. Thanks in advance and again, thanks for sharing.

[@jimpim6454]

Its because he is replacing 2^k with something he knows is smaller than it namely k^2 so obviously the equality does not hold anymore so he must write the greater than symbol.

[@michaeltheisen]

"when I was learning this stuff thousands of years ago..." the stories are true. he is a sorcerer......

[@nyashachikomwe8255]

😂😂😂I laughed hard, oh boy🤣

[@tauceti8341]

2:50 /3:20 /4:47 When dinosaurs roamed the planet xDDDD I love the humility. These are starting to click for me and it's exciting to mess with algebra like this

[@EastBurningRed]

Technically this is true for the open interval (4, infinity), so you need a more generalized induction that utilizes the well ordering relation.

[@kuldeepsharma-oc5fo]

thanks sir for the solution I was stuck on this question from last 3-4 hours. great help.from india.

[@TheMathSorcerer]

👍

[@giovannicalafiore7790]

What a smooth proof and explanation, simply wonderful, i love induction as I loved this video!!

[@EsotericArnold]

Aww man, this was beautiful, you were down to earth and showed very clearly all the things I missed from too many conversations with my professor. I actually have a good idea now, of how I should think when doing these inequality proofs. Absolutely amazing. Thank you!!

[@Shogun507]

Think I'm a bit clear now on how to approach and tackle questions which involve proving inequalities using induction. Thanks a lot for making this video!

[@israelchavez7544]

I can't say how helpful this was. I will now be ready for class tommorow. THANKS!

[@clawjet6069]

Best explanation I heard. First I thought this problem and my assignment from my pre-cal class was the same but it was actually the opposite " Prove n^2 > 2^n for n >= 5 " After watching the vid, I knew that the statement is already false so how do I show that the statement above is false using The Mathematical Induction?

[@sindbadthesailor4808]

The people want more induction proofs! Please do lots of them. (more tricky ones too)

[@TheMathSorcerer]

😄👍

[@chayajayamanna8823]

Excellent way of explaining. Night before the submission date. Thank you Sir

[@TheMathSorcerer]

You are welcome!

[@legend7890]

I took discrete math 1 year ago. I didn't understand mathematical induction. This semester I am taking theoretical CS and mathematical induction is needed so I am learning it again. This is the first time I understood a proof by Mathematical inducton.

[@ruantristancarlinsky3851]

Lol me too which University

[@TechToppers]

How I did: Checking base case is easy... I proved another inequality before that: 2^m>2m+1 (for m>4) Make hypothesis and other stuff... To proof: 2^m+2^m>2(m+1)+1 (m>4) This reduces (by hypothesis) 2^m>2 (m>4) Works! Nice! Now to the main thing: Do hypothesis and base checking... To proof 2^n+2^n>(n+1)²=n²+2n+1 This reduces to(by hypothesis): 2^n>2n+1 Proved above! So, hence proved. I suppose. Is that right? I wrote it informally... Would do better in exam... I should have gone the other way round like first write 2^k>k², add inequality I proved and then proceed. You can spare me on KZbin right?? And tell if this is right... Please? Will you marks in exam? Or in spirit of math, is the idea correct?

[@someonewhoisgreek6186]

I am a bit confused. You replaced a k with 4 (I assume because that is the lowest value it can take ). Shouldn't the domain be k greater or equal to 4 in order to use four in the proof? It works with 5 as well, I am just curious as to whether this is a simple mistake or if I don't understand something. Can someone help?

[@gunarajregmi6727]

I am also confused on it . You can't use 4 . We have to start with 5

[@ibghxr]

I think he made a mistake, it was 5 imo.

[@strugglingcollegestudent]

3:20 - said every STEM major ever.

[@callumatkinson3143]

Thanks for the vid I've been struggling with this for ages. I'm a bit confused about where you substituted in 4 for k. How does that work like would it still cover all the values bigger than 4?

[@TheMathSorcerer]

I used to struggle with your question also, tons of people do. The simple answer is that it's because k >= 4, so you can make that substitution. For example say k >= 4. And say you have 3k + p then you can write 3k + p >= 3*4 + p = 12 + p that's allowed:) You could work it out the long way. We have k >= 4, so 3k >= 12, so 3k + p >= 12 + p but nobody does that, because it's too much work. So in general, we just substitute as above.

[@CallBlofD]

Thank you for your help! Can you explain why k>=4 instead of k>4, because at start it define as k>4, how you change it to also be equal, or on what you depend when you say it. Thank you!

[@Chrisymcmb]

@@CallBlofD I was also wondering about this. The problem states that k>4, not k>=4, so that is why I was wondering how the k could be substituted by 4

[@sofiachalkias5767]

A saving grace for discrete math this semester :)) Sets theory proofs and now I found out you do induction too, LETS GOOO!!! I was wondering since we were working with k > 4 how you were able to substitute k = 4 into the equation. Because of the I.H it is totally plausible to do this but it would have to be k >= 4. Even for k>=4 this should work right? I assumed since k > 4 that we were only allowed to plug in 5 or greater for k since our I.H is greater than 4 not equal to it. Thank you!

[@sebastianohajda411]

I think because for the rule k>4, if we substitute k=4 in the LHS equation, then we know the LHS will be bigger than the substituted version of it because of the rule k>4. I think you can also you k=5, but then you have to use >= sign, since LHS can be bigger or equal to k=5 substituted version of it

[@abdoulkarenzo3138]

U are the first to teach very well me math induction thx a lot my broyher

[@TheMathSorcerer]

You are most welcome!

[@jackiesalazar2979]

Excellent way of explaining this!! It was very helpful. Thank you! 😊

[@TheMathSorcerer]

You are welcome 👍

[@doctorduggo1873]

Wow, thank you so much! Excellently explained and easy to understand after you think about it a bit.

[@RonaSue-j6c]

Thank for your explain🎉 I am studying your video from Myanmar

[@Nidhsa]

On my exam, i used induction twice for this problem. Once to prove since 2^n >n², if we can prove n²>2n +1 then 2^n + 2^n > n² + 2n +1 and the inequality is still true, and we get 2^(n+1) > (n+1)²

[@xyzpdg1313]

I am sorry, I must be dumber than the rest of the people here. Everything up to 5:00 makes sense. x + x is 2x. I nod along. Then 5:15 hits and you replace 2^k+2^k with k^2+k^2. ...wut? How did 2^k become k^2?? We had k^2 + 2k + 1, where did k^2 + k^2 come from??? Then 8:19 "replace this k with 4, gives you 8" ...ok... "replace the 8 with 1" WAT??? HOW DOES 8 BECOME 1???? I am sad. I will keep hunting.

[@shadysama3143]

How did you replace k with 4 when you're assuming for some k>4? Aren't you supposed to replace k with a number greater than 4 because its not k >= 4?

[@keis7153]

First of all, thanks for the video everything is more clear now. Today I had my first exam at college and I had to proof that if A is a countable set then so is A^n by induction. Can you make a video of that?

[@TheMathSorcerer]

Will try thank you for the idea!!

[@flidoofficial1848]

Thank you so much for doing this video, I’ve been trying to understand this for weeks

[@ANDREADELLAMAGGIORA]

Great video! The only thing I did not understand in the demonstration is why did you replace k with 4? if the hypothesis says it must be > 4 then shouldn't k be replaced with 5? Thanks a lot.

[@matko8038]

If you plug in k=5, the inequality will not hold. We want k^2+k*k to be greater than k^2+X*k. Our original assumption is that k>4 so we have to use some X that is less than k. k^2+k*k > k^2+X*k --> X4 we can use X=4.

[@EmperorKingK]

​@@matko8038Thanks for the explanation, I was really struggling to understand why he used 4. But I have another, why do we want k^2+k*k to be greater than k^2+X*k?

[@matko8038]

​@@EmperorKingKOk so, when doing induction there is a sort of rule that you aren't allowed to change one side of the equation. In this case the 2^(k+1) remains unchanged in the whole process, it is always the term on the left side. Then since we only have the right side to work with, we are trying to get the right side to create the form of some n² which is what we are trying to prove. While doing that we are free to use k>4 because that's part of our hypothesis. So to recap, we want to get (k+1)² on the right side, and 2^(k+1) must stay on the left side. Then we start working on the right side to get it into (k+1)² form, and we are starting from that first equality since we know that's true. The easiest way to get form of (k+1)² is to get form of k²+2k+1. That's why we use our hypothesis (k>4) wherever we can so we can easier get into that form, but keeping in mind if we are changing any equalities or inequalities. So finally to answer your question, It is not that we want that specifically, it's just a step that helps us get into the form we want. We already have the k² part and in this case needed the +X*k part of (k+1)²

[@EmperorKingK]

@@matko8038 I think I get it now (or at least I hope). You've been a great help either way. Thank you so much!

[@matko8038]

@@EmperorKingK You're welcome, please ask more of you get stuck on something :)

[@swatijanmeda4942]

Exact question came in my exam.... Thanks a lot.

[@TheMathSorcerer]

Great 👍

[@creamypeanutbutter6326]

Hi may i ask what property or theorem you used when you replaced 2^k to k^2?

[@foureyefreak00]

Without using “brute force”, another way of reasoning may be to compare k^2 and 2k+1. As k^2 - 2k - 1 > 0 when n > 1+sqrt(2) so k^2 > 2k+1 when n>4. This gives 2k^2 > k^+2k+1 = (k+1)^2.

[@SpartaSpartan117]

So my instinct would be to pivot once you get to the “>k^2+k^2" to proving that k^2>2k+1 for all k>3. I wonder if there is any downside to that method; specifically in how that approach of basing the proof off of another lemma may fail when it is a more difficult problem and perhaps the dependency I need is harder to prove. Any thoughts?

[@TheMathSorcerer]

that works but, it's also more work;) but yeah that could work!

[@tonyhaddad1394]

Read my comment its easy i just proove it

[@okohsamuel314]

Math Sorcerer, U sure have done well as regards providing the pathway which clearly is referring to k > 4 very insightful of u. But here is how to put it altogether. PROOF : Let's have P(n) : 2^n > n² for n > 4 Here n = 5, 6, 7, ... Base case, when n = 5 we have, P(5) : 2^5 > 5² (true) So, P(5) is true. Assume P(k) is true for some k > 5 ==> 2^k > k² ==> 2^(k+1) > 2k² ==> 2^(k+1) > k² + k² ==> 2^(k+1) > k² + 4k (k² > 4k) Better still, ==> 2^(k+1) > k² + 2k + 2k ==> 2^(k+1) > k² + 2k + 8 (2k > 8) ==> 2^(k+1) > k² + 2k + 1 (8 > 1) So that, ==> 2^(k+1) > (k+1)² ==> P(k+1) is also true. Hence, by P M I P(n) is true for all positive integers n > 4.

[@thefunnybird7246]

You are making it too complicated❗

[@okohsamuel314]

@@thefunnybird7246 ... Hahahahaha, u r actually funny ... But not 2 worry, I am very much cool with very 'COMPLICATED' math though ... so 2 me, that's not a big deal at all 🙄 ... it's just part of what really makes one a Mathematician ...which is the reason why math is usually perceived 2 be tough by people who do preconceive it 2 be 'easy' ... Math is naturally complicated for it 2 remain Math ... otherwise, it really won't be Math in the first place! 👍😂

[@thefunnybird7246]

@@okohsamuel314 the crowd of the sheep looks for simple simplification, the real ones make complex things simple, and vice-versa.

[@janetan9019]

Sir im sorry... I still don't understand why 2•2^k= (2^k)+(2^k) at 4:40

[@CG119Animator]

Since (2! or 2 = 2^{1}), we can apply the laws of exponents (n^{m} * n^{n} = n^{m+n}), we can say 2^{1} * 2^{k} = 2^{1+k} & 2^{k+1}. For (n = 0) or (n = 2), we have (n^{k}) + (n^{k}) = n^{k+1} when k is a positive integer. Therefore, 2^{k+1} = (2^{k}) + (2^{k}). Although this may not be immediately obvious, it follows directly from the properties of exponents.

[@Cokk9ine]

I’m confused

[@diamondogs0]

real

[@haleematariq4184]

Good sort of information you delivered to the viewers.

[@TheMathSorcerer]

Thx

[@FounDead]

thousand of years ago part is iconic

[@moodymonstrosity1423]

If k>4 why do we put 4????

[@oscarfc1993]

At 7:33 he has k^2 + 2k + 1. Shouldn't it be k^2 + 2k +1 + 7? If not how did he get rid of the 7?

[@mariamihab9542]

Why 7 ?

[@inquisitive871]

I see other induction inequality videos that show a different method. I find this method much more comprehensive. Would it work for all induction inequality proofs?

[@TheMathSorcerer]

Yes, absolutely, the ideas are the SAME for most of these!! thank you glad it was helpful, induction inequality is so hard to learn!!

[@ayaalkiyumix]

Thank you man . Could you please tell me the size of your whiteboard ?

[@ahsanraza135]

Oh my god this is the second time im watching this video and still not understanding it😡

[@Cokk9ine]

Ong

[@vincent2831]

at 6:24 we are claiming that (k^2) + (k^2) > (k^2) + (k*k), but shouldn't those be equal??

[@katashi1995]

Thank you for your video. K have a question... why does the 8 becomes 1 in the last part?

[@melbournebaldove6189]

Thank you very much for your videos. Do you have a good book that really tackles inequalities to have a mastery in them?

[@Uwek212]

This deserves a big fat LIKE

[@TheMathSorcerer]

Haha thx

[@jesuisravi]

Good. I've been in need of just this information.

[@TheMathSorcerer]

Glad it was helpful!

[@danielhobbyist]

First thing I thought of was proving that the equation for 2^n approaches infinity faster than n^2 using the derivative. Didnt know what induction was at the time though

[@tonyhaddad1394]

I have a simpler proof by induction : first n>4 base case : 2^5 > 5^2 32>25 true check Induction hypothesis Suppose 2^k > k^2 when k > 4 we want to show 2^(k+1) > (k+1)^2 If k> 4 then k > 1 then 3k > 2k + 1 and k^2 > 4k > 3k > 2k + 1 So k^2 + k^2 > k^2 + 2k + 1 2k^2 > (k+1)^2 2×2^k > 2k^2 (by the base case multiply both side by 2 ) then 2^(k+1) > 2k^2 > (k+1)^2 2^(k+1) > (k+1)^2 Proof done !!!!!

[@tonyhaddad1394]

****(by the induction hypothesis)

[@BayatOsman]

@@tonyhaddad1394 Thanks alot, I understand your method better.

[@arne8780]

At 4:56, I don't understand why we're "allowed" to replace 2k+2k with k^2+k^2.

[@GODCOR]

mehn.. i like the way you teach.. better than my lecturer.. lol

[@upliftingspirit6873]

So we know that k > 4 is true in the hypothesis step. In the induction step, since n = k + 1, isn't it : n > 4 => k + 1 > 4 => k > 3 ?

[@maxamedaxmedn6380]

Oh thanks I think k>3 makes sense Because i was a hard time understanding why he used k=4 In the induction step and at the same time he says k>4

[@tethyn]

Well ordering principle is important for those are learning proof by induction. Not sure if a video explaining this with these videos would be helpful or not. Other than that it is a good step by step proof with an excellent approach and thinking that is used in the types of proofs.

[@omranone1058]

I really liked this method, thank you for your effort .

[@mariamihab9542]

I don't understand why we replace K with 4 we have K is bigger than for not equal , so I don't get this point

[@musiccd1896]

Great video keep them coming. I remember i had the same assignament. Proof was for n>=3 in my case.

[@kopilahmed1887]

Here something i don't understand , here a condition that n>4 . How to you put 2×4 replacing by 4k?

[@jeroen3648]

Thank you for this tutorial, I was struck with this question, and your video helped me understand. :>

[@imnotblackpinkfanimtheirai5300]

wow, you made this problem much easier. thanks

[@ashimbasnet6568]

how do you use the same method for 4^n > n^3 for all N . I try to open it up like that and got stuck at 4^(k+1)>= k^3 +3K^2 .k

[@doodsravenlolo7173]

Thanks a lot sir, By GOD'S Grace the problem that i have now, was being solve. Keep safe and GOD Bless Always sir. Happy Mid-Week sir. And also Praise GOD sir, Praise GOD, and also to our Lord and Saviour Jesus Christ and to the Holy Spirit who is guiding as always. And To GOD Be All The Glory Always And Forever. Amen. 🙏🙏🙏🙏. Sir.

[@godwincornelius201]

So I saw the thumbnail and I was like.. wait, that's an invalid question... So I clicked on the video and am like ooh😁. N>4

[@TheMathSorcerer]

Haha yeah

[@thefunnybird7246]

2^1 * 2^k = 4^k ----- 4^k is same as 2^k +2^k 😑 Don't know if this is legal or not but it works 🤗

[@matko8038]

4^k is not the same as 2^k+2^k

[@MelvinFung]

I thought K was large than 4, so shouldn't you substitute with 5 instead?

[@zerkyl]

Where did the 2^1 gone to?

[@specialbiniyam544]

what an explanation! I really loved it sir, but, replacing "k" with 4 is not valid, as far as I'm concerned.

[@jonathanwu5245]

How did he go from +8 to +1 at the end? I still don't follow? we're suppose to set it to equal to each other?

[@jonathanwu5245]

Okay, I think it makes slightly more sense since 8 is greater than one

[@vasuhardeo1418]

very cool to split it up, and yeh is a great look at proofs

[@riyatiwari1262]

Seems like maths also have exceptions!...well thankyouu so much it was understandable.

[@MrTrollNerd]

Since the question doesn't explicitly mention only integer values of 'n', wouldn't it be more approapriate to solve it for rationals? Induction wouldn't be possible but maybe something involving the right hand limit of 4?

[@TheMathSorcerer]

it's supposed to be for integer values, oh hmm for noninteger values? I dunno I'd have to think about that one!!! Maybe what you say would work yes, not sure:) I think maybe subtracting it, and writing it like 2^x - x^2, then calling that f(x), and using some calculus, that might do it, maybe!!!

[@MrTrollNerd]

@@TheMathSorcerer I thought about it: both functions intersect at x=4, and the derivative of the 2^x term is always greater than that of the x^2 term for x>4. So it becomes trivial, I suppose. It probably doesn't make sense to make it more formal

[@rhodanyangu168]

This is amazing, I was given the first question to work out. Thanks 😍

[@farhansyed1038]

why can you replace k with 4?

[@yaloow]

Wait so how did the 8 turn into one

[@nathanyao3525]

Thank you so much. After I see the solution to a proof question that I don't know how to do, I'm always wondering to myself, "how the heck was I supposed to know to do that?" Do you have any tips?