Prove ##5^n+9<6^n## for ##n\epsilon N|n\ge2## by induction

AI Thread Summary
The discussion revolves around proving the inequality \(5^n + 9 < 6^n\) for \(n \in \mathbb{N}\) where \(n \ge 2\) using mathematical induction. The base case for \(n=2\) is verified as \(34 < 36\). The induction hypothesis assumes the inequality holds for \(n=k\), and the induction step demonstrates that if \(5^k + 9 < 6^k\), then \(5^{k+1} + 9 < 6^{k+1}\) follows by multiplying both sides by appropriate constants. Participants emphasize clarity in presentation and logical flow in proofs, suggesting a structured approach to writing mathematical arguments. Overall, the proof is validated, with recommendations for improving clarity and mathematical rigor.
McFluffy
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Homework Statement


Prove ##5^n+9<6^n## for ##n\epsilon \mathbb{N}|n\ge2## by induction.

Homework Equations


None

The Attempt at a Solution


The base case which is when ##n=2##:

##5^2+9<6^2##
##34<36##

Thus, the base case is true. Now for the induction step.

Induction hypothesis: Assume ##5^k+9<6^k## for ##k \epsilon \mathbb{N}|k\ge2##

Induction step: To show that ##5^k+9<6^k## implies ##5^{k+1}+9<6^{k+1}##, I'll start with the induction hypothesis:

##5^k+9<6^k##
##5(5^k+9)<6(6^k)## since we have an inequality and not an equation, I can add numbers that are bigger to one side and not violate inequality. With this in mind, since ##5<6##, I'd multiplied 5 on the smaller side and 6 on the bigger side.

##5^{k+1}+45<6^{k+1}##
##5^{k+1}+9<6^{k+1}##

Multiply one side by a bigger number than the other side is what makes this proof possible. And I think this is a logical thing to do because we're dealing with an inequality and not an equation. Are my steps justified?
 
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McFluffy said:

Homework Statement


Prove ##5^n+9<6^n## for ##n\epsilon \mathbb{N}|n\ge2## by induction.

Homework Equations


None

The Attempt at a Solution


The base case which is when ##n=2##:

##5^2+9<6^2##
##34<36##

Thus, the base case is true. Now for the induction step.

Induction hypothesis: Assume ##5^k+9<6^k## for ##k \epsilon \mathbb{N}|k\ge2##

Induction step: To show that ##5^k+9<6^k## implies ##5^{k+1}+9<6^{k+1}##, I'll start with the induction hypothesis:

##5^k+9<6^k##
##5(5^k+9)<6(6^k)## since we have an inequality and not an equation, I can add numbers that are bigger to one side and not violate inequality. With this in mind, since ##5<6##, I'd multiplied 5 on the smaller side and 6 on the bigger side.

##5^{k+1}+45<6^{k+1}##
##5^{k+1}+9<6^{k+1}##

Multiply one side by a bigger number than the other side is what makes this proof possible. And I think this is a logical thing to do because we're dealing with an inequality and not an equation. Are my steps justified?
Yes. Only your presentation could be improved. E.g.
##5^{k+1} + 9 < 5^{k+1}+45 = 5\cdot (5^k+9) <_{I.H.} 5 \cdot 6^k < 6 \cdot 6^k = 6^{k+1}##
 
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McFluffy said:

Homework Statement


Prove ##5^n+9<6^n## for ##n\epsilon \mathbb{N}|n\ge2## by induction.

Homework Equations


None

The Attempt at a Solution


The base case which is when ##n=2##:

##5^2+9<6^2##
##34<36##

Thus, the base case is true. Now for the induction step.

Induction hypothesis: Assume ##5^k+9<6^k## for ##k \epsilon \mathbb{N}|k\ge2##

Induction step: To show that ##5^k+9<6^k## implies ##5^{k+1}+9<6^{k+1}##, I'll start with the induction hypothesis:

##5^k+9<6^k##
##5(5^k+9)<6(6^k)## since we have an inequality and not an equation, I can add numbers that are bigger to one side and not violate inequality. With this in mind, since ##5<6##, I'd multiplied 5 on the smaller side and 6 on the bigger side.

##5^{k+1}+45<6^{k+1}##
##5^{k+1}+9<6^{k+1}##

Multiply one side by a bigger number than the other side is what makes this proof possible. And I think this is a logical thing to do because we're dealing with an inequality and not an equation. Are my steps justified?

Yes it is all correct. Multiplication here is just adding and you can always add smaller term to a side and larger term to other side without flipping the sign.

Also you may want to use ##\in## (\in) instead of ##\epsilon##.

Sorry I did not see the post by @fresh_42 , it did not get updated on page.
 
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McFluffy said:
Multiply one side by a bigger number than the other side is what makes this proof possible. And I think this is a logical thing to do because we're dealing with an inequality and not an equation. Are my steps justified?

Yes, but you can show it mathematically, with less words: ##5(5^k+9)<5(6^k)<6^{k+1}##

Also, you can write in the last step ##5^{k+1}+9<5^{k+1}+45<6^{k+1}##
 
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fresh_42 said:
Yes. Only your presentation could be improved. E.g.
##5^{k+1} + 9 < 5^{k+1}+45 = 5\cdot (5^k+9) <_{I.H.} 5 \cdot 6^k < 6 \cdot 6^k = 6^{k+1}##
ehild said:
Yes, but you can show it mathematically, with less words: ##5(5^k+9)<5(6^k)<6^{k+1}##

Also, you can write in the last step ##5^{k+1}+9<5^{k+1}+45<6^{k+1}##

What are the general guidelines that I should follow when writing a proof like this? I'm still quite new to proofs.
 
McFluffy said:
What are the general guidelines that I should follow when writing a proof like this? I'm still quite new to proofs.
Good question. I think the important part is to always make clear what follows from what. Start with what you have, in this case ##5^k+9<6^k## and either ##5^{k+1}+9## and make it bigger, or ##6^{k+1}## and make it smaller. Personally I find it easier to read when it get's bigger, but this is a matter of taste. If you write your expressions from one line to the next and always handle both sides, it is more difficult to see whether you follow a path or simply write down some expressions. If both sides ##5^{k+1}+9## and ##6^{k+1}## appear in the same line, then it's hard to tell, whether you still deal with the claim or already entered the proof.
Of course you can do this in a draft and order the terms afterwards for the final version. Imagine if it were you to read it. A proof should lead from what is given to what has to be shown and the steps in between should show the implications needed. As I said, it wasn't wrong what you've written, only a bit confusing to read. You have @ehild's and mine example. Compare them with yours.

As I learned it, I followed a short template.

Condition: Given facts.
Statement: What has to be shown.
Proof: ... and in case it was an indirect proof, I wrote. Assumption: (the opposite of what has to be shown)
Now come the steps ##\Rightarrow## ... ##\Rightarrow## ...
And at the end a sign that it is the end, like a small box or so. And sure, I abbreviated these underlined words.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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