Comparing Almost 1s
Which of these numbers is larger?
2 0 1 5 2 0 1 6 , 2 0 1 6 2 0 1 7
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9 solutions
Moderator note:
Good way to simplify the inequality.
Nicely done!
For any positive integer n > 1 consider this:
n 2 > n 2 − 1
⇒ n × n > ( n + 1 ) × ( n − 1 )
⇒ n − 1 n > n n + 1
Substituting n=2016 :
⇒ 2 0 1 5 2 0 1 6 > 2 0 1 6 2 0 1 7
(+1).. Did the same way..
Actually, in any real number, it is true - even zero or a negative. Since it is a squared expression, a negative or zero number ,in that expression, when squared, is greater than the right side.
Note that n can be 1.
EDIT: I said that only for the first statement.
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n-1 is coming in denominator which restricts n from taking the value 1!!
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@Mehul Arora – Get a room guyzz :P very nice explanation though
Don't get me wrong, I accidentally clicked unfollow :P
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@Mehul Arora
–
Yeah .. I was terrified- how a person could follow someone twice!! ...
Anyways its irrelevant but why you dislike WWE .... (If you could answer, no force from my side)..
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- It's unreal. Really unreal.
What's the point of seeing two random people beat each other up?
The signature moves and stuff are lame. Kids try to copy it all the time, and it is lame as anything.
They fool people too much.
P.S. How are you always online?
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@Mehul Arora – I use android for surfing brilliant and always receive a notification from my mail id when someone replies.... Anyways (+1) for a nice explanation.. ;-)
\begin{aligned} \color{#3D99F6}{\dfrac {1}{2015}} \quad \boxed {>}& \quad \color{#D61F06}{\dfrac {1}{2016}} \\ \color{#3D99F6}{\dfrac {1}{2015}+1} \quad \boxed {>}& \quad \color{#D61F06}{\dfrac {1}{2016}+1} \\\\ \therefore \color{#3D99F6}{\dfrac{2016}{2015}}\quad \boxed {>}& \quad \color{#D61F06}{\dfrac{2017}{2016}} \end{aligned} \end {equation}
To compare fractions: A/B versus C/D
Cross multiply: A D and B C
Larger value is the side where the larger fraction is.
2016X2016 = 4064256 and 2015X2017 = 4064255
so 2016/2015 is larger
Let n = 2 0 1 5 , n + 1 = 2 0 1 6 , n + 2 = 2 0 1 7
2 0 1 5 2 0 1 6
= n n + 1
= n n + n 1
= 1 + n 1
2 0 1 6 2 0 1 7
= n + 1 n + 2
= n + 1 n + 1 + n + 1 1
= 1 + n + 1 1
Now, we know that:
n < n + 1
n 1 > n + 1 1
1 + n 1 > 1 + n + 1 1
Therefore, the general rule for these kind of fractions is:
n n + 1 > n + 1 n + 2
For our specific case here:
2 0 1 5 2 0 1 6 > 2 0 1 6 2 0 1 7
Note: The general rule also works for negative numbers. Test it out!
Note note: This general rule does not work for n = 0 or n = − 1
Consider a function x+1/x it is a decreasing function as its derivative is negative for all real x so lower value of x higher the value of function
(2016\2015)÷(2017\2016)=2016^2÷{2015×2017)=2.016^2÷{2.015×2.017}. (2+.016)^2=4+.064+.000256={4.064256} (2+.015)×(2+.017)=4+.034+.030+.000255. ={4.064255} so {2016 ÷2015}÷{2017÷2016}>1 so (2016÷2015)>(2017÷2016)####
2 0 1 5 2 0 1 6 ? 2 0 1 6 2 0 1 7
1 + 2 0 1 5 1 ? 1 + 2 0 1 6 1
2 0 1 5 1 ? 2 0 1 6 1
1 ? 2 0 1 6 2 0 1 5
1 > 2 0 1 6 2 0 1 5
2 0 1 5 1 > 2 0 1 6 1
1 + 2 0 1 5 1 > 1 + 2 0 1 6 1
2 0 1 5 2 0 1 6 > 2 0 1 6 2 0 1 7
If a>b>c then, (b/c)>(a/b) Here, a =2017 & b=2016 & c = 2015
if a > b > c
you say c b > b a
let a = 6 , b = 2 , c = 1 and substitute:
1 2 > 2 6 = 3
the inequality wrong.
2 0 1 5 1 > 2 0 1 6 1 ⇒ 1 + 2 0 1 5 1 > 1 + 2 0 1 6 1 ⇒ 2 0 1 5 2 0 1 5 + 2 0 1 5 1 > 2 0 1 6 2 0 1 6 + 2 0 1 6 1 ⇒ 2 0 1 5 2 0 1 5 + 1 > 2 0 1 6 2 0 1 6 + 1 ⇒ 2 0 1 5 2 0 1 6 > 2 0 1 6 2 0 1 7