Correct me if I am wrong

  • #1

Main Question or Discussion Point

Prove that of 25 is subtrated from the square of an odd integer greater than 5, the resulting number is always divisible by 8.

Solution:

Let S (n) = n2 – 25
For n = 1, S (1) = (1 )2 – 25 = -24 which is clearly divisible by 8.
Thus the first condition is satisfied as S(1) is true.

Let us assume that S(n) is true for n = 2k+1 Belonging to Odd integers greater than 5, that is,

S(2k+1) = (2k+1)2 - 25 is divisble by 8
= 4k2 + 4k + 1 -25
= 4k2 + 4k - 24
= 4(k2 + k - 6)
= 4(k-2)(k+3) ------ (A)

We can see that (A) is clearly divisible by 8, condition being that k is an odd integer which is bigger than 5. Since both conditions are satisfied, hence by mathematical induction we have proven that S(n) is divisble by 8 for all integers bigger than 5.
 

Answers and Replies

  • #2
379
0
umm, you are not exactly correct. you are setting n = to 2k + 1. while that is denoting an odd integer, that is what n should be and you have not shows for n+1 yet which should be 2k + 3.

however, induction doe snot require that you use that form. you can simply say n for all n = 2k+1, now showing for n+1 and just use n+1 to show it rather than 2k+3
 
  • #3
shmoe
Science Advisor
Homework Helper
1,992
1
Jarvis Bull Dawg said:
Let S (n) = n2 – 25
For n = 1, S (1) = (1 )2 – 25 = -24 which is clearly divisible by 8.
Thus the first condition is satisfied as S(1) is true.
Why is your base case n=1 when you're trying to prove this for odd integers greater than 5?


Jarvis Bull Dawg said:
Let us assume that S(n) is true for n = 2k+1 Belonging to Odd integers greater than 5, that is,

S(2k+1) = (2k+1)2 - 25 is divisble by 8
= 4k2 + 4k + 1 -25
= 4k2 + 4k - 24
= 4(k2 + k - 6)
= 4(k-2)(k+3) ------ (A)
You don't use your induction hypothesis here, and you don't need to. Induction is not necessary.

Jarvis Bull Dawg said:
We can see that (A) is clearly divisible by 8, condition being that k is an odd integer which is bigger than 5.
n>5, not k.
 
  • #4
shmoe said:
Why is your base case n=1 when you're trying to prove this for odd integers greater than 5?
True. Thank you :smile:




shmoe said:
You don't use your induction hypothesis here, and you don't need to. Induction is not necessary.

What should I use?

Next thing, so should I use k+1 rather than 2k+1? Advice?
 
  • #5
shmoe
Science Advisor
Homework Helper
1,992
1
Jarvis Bull Dawg said:
What should I use?

Next thing, so should I use k+1 rather than 2k+1? Advice?
2k+1 is good, n is an odd integer after all. You don't know if k is even or odd though.

use what you've got:

n^2-25
=(2k+1)^2 - 25
= 4k^2 + 4k + 1 -25
= 4k^2 + 4k - 24
= 4(k^2 + k - 6)
= 4(k-2)(k+3) ------ (A)

Show A is divisible by 8. 2 cases if you like, k even or k odd.

Actually the n>5 bit isn't necessary, it's just to keep n^2-25 positive, the concept of divisibility still makes sense for negative integers.

By the way you can use Latex to make exponents [tex]n^{2}[/tex] <-click to see how, or please use a ^ to denote exponentiation as I've done above, as "n2" looks like another variable or the variable n times 2. :smile:
 
  • #6
rachmaninoff
Odd integers greater than 5: 7, 9, 11...
(5+2), (5+4), (5+6)...

general form: (5+2n), n=1,2,3...


So, what number are you looking at? (5+2n)^2 minus something? Do some algebra here...

Until you're done. (hint: factor out '2's when you can...)

There's no need for induction. If you must, start with n=7, not n=1 (factoring negative integers is uncool).
 
Last edited by a moderator:
  • #7
28
0
I think it is easier solved this way

(n^2)-25 divides 8 is equivalent to

(n^2)-1 divides 8 because 24 divides 8

now we have

(n-1) (n+1) divides 8

n= 2 m + 1 ( n is odd as suggested in the problem)

(2 m ) (2 m + 2 ) = 4 m (m+1)

if m is even we are done , if m is odd then (m+1) is even .Both cases prove

divisibility by 8
 
  • #8
Here is another one, but seriously I have no idea what this is.

Observe that the last two digits of [tex]7^2[/tex] are 49, the last two digits of [tex]7^3[/tex] are 43, the last two digits of [tex]7^4[/tex] are 01, and the last two digits of [tex]7^5[/tex] are 07. Prove that the last two digits of [tex]7^201[/tex] are 07.
 
  • #9
shmoe
Science Advisor
Homework Helper
1,992
1
what are the last two digits of [tex]7^6[/tex]? [tex]7^7[/tex]? Notice a pattern?

Suggestion:[tex]7^{4}=100k+1[/tex] for some integer k. Use this to show that the last two digits of [tex]7^{4n}[/tex] are 01 for any natural number n.
 
  • #10
shmoe said:
what are the last two digits of [tex]7^6[/tex]? [tex]7^7[/tex]? Notice a pattern?

Suggestion:[tex]7^{4}=100k+1[/tex] for some integer k. Use this to show that the last two digits of [tex]7^{4n}[/tex] are 01 for any natural number n.

Yes I do notice the recurring pattern, but I didn't get your suggestion.
 
  • #11
shmoe
Science Advisor
Homework Helper
1,992
1
Knowing [tex]7^{4}=100k+1[/tex] you should be able to prove you can write [tex]7^{4n}=100m+1[/tex] where m is some integer (for example use the binomial theorem).

This will prove that the last two digits of [tex]7^{200}[/tex] are 01. (Since you're only interested in this case you could really take n=50 above, but it doesn't really simplify things)

By the way, you're not familiar with modular arithmetic are you? While not necessary to do problems like this, it does simplify things.
 
  • #12
Edited for errors
 
Last edited:
  • #13
Hope yo get it. I solved it for the 201 term. By B-Theorem
 
  • #14
28
0
Jarvis Bull Dawg said:
Here is another one, but seriously I have no idea what this is.

Observe that the last two digits of [tex]7^2[/tex] are 49, the last two digits of [tex]7^3[/tex] are 43, the last two digits of [tex]7^4[/tex] are 01, and the last two digits of [tex]7^5[/tex] are 07. Prove that the last two digits of [tex]7^201[/tex] are 07.
I tried to do it this way
By binomial theorem I find the last two terms are
(10-3) ^ 201 = (201) (10) (3 ^ 200 )- (3) ^ 201
or
(3 ^ 200) (2010 - 3)
or
(9^100) (2000+7) =((10-1)^100 )( 7) since we are concerned with the last two digits
or
7(1000 t +1)
7000 t +7
This proves the last two digits are 07
 
  • #15
1,425
1
You're solutions are perhaps correct but you fail to proove the generality of this statement (for any therm). The true solution is much simpler...

Any odd number greater than 5 can be writtent under the form 5+x, where x is even. Following the statement:

(5 + x)^2 - 25

25 + x^2 + 2(5)x - 25

x^2 + 2(5)x.

Since x is even, let x=2y

(2y)^2 + 2(2)(5)y

4y^2 + 4(5y)

4(y^2 + 5y)

y^2 + 5y is even for any integrer. So let y^2 + 5y=2a

4(2a)

8a

This is the proof that the solution for (5 + x)^2 - 25, where x is even is always a factor of 8.
 
Last edited:

Related Threads for: Correct me if I am wrong

Replies
35
Views
6K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
5
Views
2K
Top