Prove that successive differences between square roots of integers decreases without using limits or

Matthew Hubbard

Matthew Hubbard

Answered question

2022-05-13

Prove that successive differences between square roots of integers decreases without using limits or derivatives
I have been trying to prove that n + 1 n > n + 2 n + 1 for ALL n = 1 , 2 , 3 , . . . without success.

Answer & Explanation

Jaylon Richmond

Jaylon Richmond

Beginner2022-05-14Added 10 answers

Step 1
The answers in the link are sufficient, since the identity n + 1 n = 1 n + 1 + n immediately implies n + 2 n + 1 = 1 n + 2 + n + 1 ,, and since n + 2 > n ,
Step 2
it follows that n + 1 n = 1 n + 1 + n > 1 n + 2 + n + 1 = n + 2 n + 1 .
I don't know why you think this is not rigorous.
Amappyaccon22j7e

Amappyaccon22j7e

Beginner2022-05-15Added 3 answers

Explanation:
( n + 2 + n ) 2 = 2 n + 2 + 2 n 2 + 2 n < 2 n + 2 + 2 n 2 + 2 n + 1 = 4 n + 4 = ( 2 n + 1 ) 2
n + 2 + n < 2 n + 1
Some steps are left, but they are obvious.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Discrete math

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?