MHB Solve a Problem & Impress Friends: Proving a/(b+c) for Integers a, b, c

  • Thread starter Thread starter CONRADDODD
  • Start date Start date
AI Thread Summary
The discussion centers on proving that if an integer a divides both integers b and c, then a also divides the sum b + c. The proof begins by defining that if a divides b, there exists an integer k1 such that b = ak1, and similarly for c with k2. By combining these, it shows that b + c = a(k1 + k2), indicating that a divides b + c. The intuitive explanation emphasizes that since b and c are multiples of a, their sum must also be a multiple of a. This proof effectively demonstrates the divisibility property, leading to an impressive conclusion.
CONRADDODD
Messages
5
Reaction score
0
I'd love to solve this problem and stun a friend...
Prove if a/b and a/c then a/(b+C)
let a,b,c (element of) (integers.)

I'm sure there's a genius that can figure this out.

Thanks!
 
Mathematics news on Phys.org
I take it that "a/b" means "a divides b". Then, if $a$ divides $b$, then there exists an integer $k_1$ such that $b = ak_1$. Similarly, if $a$ divides $c$, then there exists an integer $k_2$ such that $c = ak_2$. Then $b + c = ak_1 + ak_2 = a(k_1 + k_2) = ak_3$ for $k_3 = k_1 + k_2$, and so $a$ divides $b + c$.

More intuitively, if $a$ divides both $b$ and $c$, that means that $b$ and $c$ are both multiples of $a$ (by definition). So their sum must also be a multiple of $a$, and the result follows.​
 
Thanks for the help!
You should have seen the look on his face! Priceless!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top