×

Awesome proof

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

Note by Shivamani Patil
2 years, 11 months ago

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 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Comments

Sort by:

Top Newest

(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 $$k^{2}$$ to $$(k+1)^{2}-1$$ are not inside, so all numbers from $$k^{4}$$ to $$(k+1)^{4}-1$$ are not inside, ...

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

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

- 2 years, 11 months ago

Log in to reply

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

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

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

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

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

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

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

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

- 2 years, 11 months ago

Log in to reply

Nihar Mahajan,Sharky Kesa,Satvik any one?

- 2 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...