Division algorithm -Doubled ,tripled

I can't prove it.Can you?

#NumberTheory #DivisionAlgorithm #ProofTechniques

Note by Shivamani Patil
6 years, 6 months 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

Division Algorithm states that a=q1b+r1a=q_{1}b+r_{1} where r1<b\leq r_{1} < b so we need to have r=r1+2br=r_{1} + 2b to get the wanted inequality. Now substituting r1r_{1} in terms of rr; a=q1b+r2ba=q_{1}b+r-2b with which we can get the value of qq; a=(q12)b+ra=(q_{1}-2)b+r Hence, q=q12q=q_{1}-2, satisfying the equation a=qb+ra=qb+r

Marc Vince Casimiro - 6 years, 6 months ago

Log in to reply

Use \leq for \leq . Nice proof.

Siddhartha Srivastava - 6 years, 6 months ago

Log in to reply

Thanks! Had been trying to find that LaTeX.

Marc Vince Casimiro - 6 years, 6 months ago

Log in to reply

@Marc Vince Casimiro With the new display Latex tools, you just need to find some problem / note / solution with the corresponding latex that you need :)

Calvin Lin Staff - 6 years, 6 months ago

qb+2b<=qb+r<qb+3b , b(q+2)<=qb+r<b(q+3) , b(q+2)<=a<b(q+3),divide by b>0, q+2<=a/b<q+3, This means that rational number a/b is between two integers...

Nikola Djuric - 6 years, 6 months ago
×

Problem Loading...

Note Loading...

Set Loading...