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Math Contest #27 Results and Solution by quantumdeveloper

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· @quantumdeveloper ·
$0.28
Math Contest #27 Results and Solution
## Solution
The problem of this contest was to proof an equivalence:
![](https://steemitimages.com/p/NTy4GV6ooFRkjTArCrebYc2WCCmX2KY4SfTbUDpHWg6ZYGGDQxkSJMEeDstbGRGui2ebEURsduenEkUUxQ1WAp1yvfxKEXKyyX4zMKofsf5yNW7Pd9n6zTpc6CpRwT9kgr9wTfq9tgpfHYz7DKGgQ1sUYSTA6e52k49nouuG?format=match&mode=fit&width=640)
I also just noticed that this inequation only holds for uneven, positive values of n(which is probably the reason why I had no entries this week).
First of all a small explanation why this is even true:
if x is bigger than 0: Then 1/x^n is only < 1 if x > 1.
if x is smaller 0: Then 1/x^n is only < -1 if x > -1
These two cases connect both inequations.
In this case you can rearrange the leftern inequation to:
`1 < x^(n+1) if x > 0` → `x > 1`
`1 > x^(n+1) if x < 0` → `1 > ±x` → `-1 < ±x` → `-1 < x`
The ± comes from taking an even root.
So if x > 0 than x > 1 is the only solution and if x < 0, x > -1 is the only solution.
Putting the 2 together gets the rightern side of the equivalence:
`x > 1` or `-1 < x < 0`
And the way back to prove it actually is an equivalence:
`x > 1` → `x^(n+1) > 1` → `x > 1/x^n`
`-1 < x < 0` → `1 < x^(n+1)` → `x > 1/x^n`

*↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓*
### List of participants with their entries:
Name | solution found | comment
 - | - | -
@tonimontana | none | found one of the errors I made → you get the reward

*↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓*

## Winner draw:
Not necessary since only 1 participant:
Congratulations @tonimontana , you won 2 SBI!
*↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓*
### The next contest starts soon. Don't miss it!
### This contest will continue as a 2-week contest as it's getting harder to find a unique problem and I have more stress lately.
👍  , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,
👎  , , , , , , ,
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