Question

Demonstrate Goldbach's conjecture for even numbers between 100-120.

          Demonstrate Goldbach's conjecture for even numbers between 100-120.
        
Demonstrate Goldbach's conjecture for even numbers between 100-120.

Added by Laura J.

Close

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Demonstrate Goldbach's conjecture for even numbers between 100-120
Close icon
Play audio
Feedback
Powered by NumerAI
Ivan Kochetkov Kathleen Carty
Danielle Fairburn verified

Jan-Luka Fatras and 85 other subject Calculus 3 educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
formulate-a-conjecture-about-the-final-two-decimal-digits-of-the-square-of-an-integer-prove-your-con

Formulate a conjecture about the final two decimal digits of the square of an integer. Prove your conjecture using a proof by cases.

Discrete Mathematics and its Applications

The Foundations: Logic and Proofs

Proof Methods and Strategy

number-sense-goldbachs-conjecture-states-that-every-even-integer-greater-than-2-can-be-expressed-as-

Number Sense Goldbach's conjecture states that every even integer greater than 2 can be expressed as the sum of two prime numbers. Here are some examples. $$4=2+2 \qquad 6=3+3 \qquad 8=3+5$$ a. Test Goldbach's conjecture on $10,12,$ and $100 .$ b. Does the conjecture seem to be true? c. What would you need to show in order to prove the conjecture?

Impact Mathematics Algebra and More

Quadratic and Inverse Relationships

Conjectures

show-there-are-infinitely-many-positive-integers-such-that-n1-is-divisible-by-at-least-two-distinct-primes-67033

'Show there are infinitely many positive integers / such that n!+1 is divisible by at least two distinct primes.'

Haricharan G.


*

Recommended Textbooks

-
Calculus: Early Transcendentals

Calculus: Early Transcendentals

James Stewart 8th Edition
achievement 1,875 solutions
Calculus: Early Transcendentals

Calculus: Early Transcendentals

William Briggs, Lyle Cochran, Bernard Gillet 3rd Edition
achievement 1,187 solutions
Thomas Calculus

Thomas Calculus

George B. Thomas Jr. 14th Edition
achievement 1,614 solutions

*

Transcript

-
00:01 All right, so first of all, notice that when you've got a number n, you can always write it as n equals 100a plus b.
00:09 I mean, a can even take the value zero for the first values, but you can always write it in this form here.
00:15 The reason i'm saying this is that n squared, so is equal to 100a plus b squared, which is 10 ,000 a squared plus 200 ab, plus b squared.
00:33 Now notice that this number here will always have the last two digits being equal to zero zero, all right? because you've got this 10 ,000 here.
00:44 And now here again, since you've got 200, the last two digits of this product here will also be zero zero.
00:52 So what we've just shown here is that you only have to consider the the b basically of the if you decompose n in this shape where let me just precise that b is between 0 and 99 well by checking only the last two digits of n well you can know the last two digits of n squared and that's already a big step because if we're trying to show something for all n we're basically trying to show it for an infinite number of values um which is is not something possible in if you want to prove something by exhaustion or proof something by being doing a proof by case and now we've reduced it to 100 different cases so one way of doing it would be simply to check for all the different values of b b ranging from 0 to 99 of b square sorry where b ranges from 0 to 99 and conclude but that is obviously very uh takes of time um and it's quite boring.
02:04 So what you want to do is to reduce the number of cases to check.
02:10 And by further, you want to further reduce the number of cases to check.
02:17 Now, why exactly did i straight, jumped straight into this decomposition? well, intuitively, you got to tell yourself that, you know, if you've got a number like 6803 ,000, 5 ,000.
02:34 440, that this part here, when it is squared, it doesn't really contribute that much to the last two digits of the number when it's squared.
02:48 Because these, i mean, this is a fuzzy argument again, but this is often what leads you to some form of more rigorous proof.
02:58 Okay.
03:00 All right.
03:00 So that was a short parenthesis.
03:03 So how do you reduce the number of cases? to check once you've established that you only have to look at b ranging from 0 to 99.
03:11 Well, the first thing is to notice that 100 minus b squared, right, is equal to 100 squared minus 200b plus b squared.
03:28 And since you only care about the last two digits, well, again, you notice that this ends in 0, this ends again in zero zero by the same argument we've made before and therefore the last two digits of 100 minus b squared is the same as the last two digits of b squared so what does that tell us well basically we've reduced the number of cases to check by two we've divided by the number of cases by two why? well, basically what this means is that checking the last two digits of a number, let's say four, is the same as a four square, sorry, is the same as checking the last two digits of 100 minus four squared, which is 96.
04:18 All right? 96 squared, sorry.
04:22 And, you know, if you check this, four squared is equal to 16 and 96 squared.
04:28 Is 9 ,260.
04:32 So you know, you see that both end in 60.
04:35 Okay.
04:36 And why do i say 50? well, i mean, you can see that.
04:40 If, you know, you start with b equals 1, then that tells you that 99 has the same, 99 squared has the same last digits as 1 squared.
04:55 Then you'd say 2, same as 98, and you go down, go down, go down, and you've basically divided, oh sorry, you've divided those 100 numbers in two halves where you've got 50 on both sides.
05:09 Okay.
05:10 Now that's already very good.
05:12 You divide the number of case by two, but you can go one step further.
05:16 And the thing is here, if you notice here, that two is, sorry, that two here is almost too big.
05:28 It's not necessary.
05:28 You could even, it could even work if that 200 was 100.
05:35 So what we do is we actually look, okay, well, what happens if i get 50 plus b squared, all right? and 50 minus b squared.
05:46 The reason i'm taking 50 is that two times 50 is 100...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever