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# Proofs

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please explain, thank you very much!

Nov 17, 2018

#1
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11) $$\frac{α}{β}> \frac{2α-β}{α+4β}$$ multiply by α+4β and β and its >0 because α,β its positive numbers

$$α(α+4β)>β(2α-β) <=> α^2+4αβ > 2αβ-β^2 <=> α^2 +4αβ - 2αβ + β^2 >0 <=> α^2 +2αβ + β^2 >0 <=> (α+β)^2>0$$

where applicable! So we prove it!

12) Something is wrong here or I don't understand the question:

exp.:$$3^2-2^2=9-4=5$$  But there is no sum of two consecutive odd numbers be equal with 5 because 1+3 = 4 and general  sum of two consecutive odd numbers  is allways be equal with even number because if n is a number we have:

n+(n+2)=2n+2 = 2(n+1) for every n natural so we have 2(n+1)

is divided by 2 so it's even number and 5(in this example) it's not even number.

Hope this helps!

Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
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Here's  12

Assuming that  we are dealing with positive integers....let the odd numbers  be

2n - 1   and 2n + 1

Adding these together we get   4n

Sometimes.....in these kinds of proofs, it is good to see if we can discover a pattern....

Let n = 2

And the sum of the odds =  8

And the difference of two squares that equals this  is 3^2 - 1^2 = 9 - 1 = 8

We can write this as  [ 2n - ]^2 - [ 2n - 3]^2

Note that the numbers in red are just  (n - 1) and (n + 1), respectively

Now......Let  n = 3

And the sum of the odds = 12

And the difference of two squares that equals this is 4^2 - 2^2  = 16 - 4  = 12

Which we can write as [ 2n - ]^2 - [ 2n - 4 ]^2

Note that, once again, the numbers in red are just ( n - 1)  and  ( n + 1), respectively

So....it appears that we can write the larger  square  as   [2n - (n- 1)]^2   =  [ n + 1]^2

And it appears that we can write the other square as  [ 2n - (n + 1) ]^2  =  [ n - 1]^2

And their difference is

[ n + 1]^2  -   [ n - 1]^2   =

[ n^2 + 2n + 1 ]  - [ n^2 - 2n + 1]  =

4n

Nov 18, 2018
edited by CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018