Awesome proof

Let MM be a nonempty set of positive integers such that 4x4x and [x]\left[ \sqrt { x } \right] both belong to MM whenever xx does.Prove that MM is the set of all natural numbers.

#NumberTheory #MathematicalInduction #Sets #Proofs

Note by Shivamani Patil
6 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

(1) If we take the floor of square roots repeatedly we will end up with 1. Hence 1 is in M.

(2) 4 and 2 are in M.

(3) All powers of 2 are in M.

(4) Let's say k is not inside, then all numbers from k2k^{2} to (k+1)21(k+1)^{2}-1 are not inside, so all numbers from k4k^{4} to (k+1)41(k+1)^{4}-1 are not inside, ...

Hence for all natural nn the numbers from k2nk^{2^{n}} to (k+1)2n1(k+1)^{2^{n}}-1 are not inside.

(5) Choose nn large enough such that the ratio between these two values is way greater than 2. Contradiction!

Joel Tan - 6 years, 1 month ago

Log in to reply

Joel Tan good one.I have similar thing.For your 5th step I have something..

f(x)=log2xf\left( x \right) =\log _{ 2 }{ x } defined for R+R{ R }_{ + }\rightarrow R is increasing and hence we have

log2(n+1)log2n>0\log _{ 2 }{ (n+1) } -\log _{ 2 }{ n } >0

Since g(x)=2xg\left( x \right) ={ 2 }^{ -x } is decreasing,for a sufficiently large positive integer hh we will have

2h<log2(n+1)log2n{ 2 }^{ -h }<\log _{ 2 }{ (n+1) } -\log _{ 2 }{ n } \quad

Which implies (n+1)2h>2n2h{ (n+1) }^{ { 2 }^{ h } }>{ 2n }^{ { 2 }^{ h } }

Therefore interval [n2h,2n2h][{ n }^{ { 2 }^{ h } },{ 2n }^{ { 2 }^{ h } }]\quad is totally contained in [n2h,(n+1)2h)[{ n }^{ { 2 }^{ h } },{ (n+1) }^{ { 2 }^{ h } })\quad .

But [n2h,2n2h][{ n }^{ { 2 }^{ h } },{ 2n }^{ { 2 }^{ h } }]\quad contains a power of 22.A contradiction.

shivamani patil - 6 years, 1 month ago

Nihar Mahajan,Sharky Kesa,Satvik any one?

shivamani patil - 6 years, 1 month ago
×

Problem Loading...

Note Loading...

Set Loading...