Prove that if a & b are odd then a+b is even

  • Thread starter Thread starter sonadoramante
  • Start date Start date
  • Tags Tags
    even
AI Thread Summary
If a and b are odd integers, then their sum a+b is even. The proof begins by expressing a and b in the form of 2n+1 and 2m+1, respectively. Adding these gives a+b = 2n + 2m + 2, which can be factored to 2(n+m+1), confirming that the sum is even. A suggestion was made to clarify the justification of why 2(n+m+1) is even. The discussion emphasizes the importance of clear mathematical reasoning in proofs.
sonadoramante
Messages
19
Reaction score
3
Summary:: Prove that if a is an odd integer and b is an odd integer then a+b is even.

Theorem: If a is odd and b is odd then a+b is even.

Proof: Let a and b be positive odd integers of the form a = 2n+1 & b = 2m+1

a+b = 2n+1+2m+1
= 2n+2m+1+1
= 2n+2m+2
= 2(n+m)+2
= Let k = n+m
= 2k+2
Therefore a+b is even.
 
Physics news on Phys.org
This mostly looks fine, though you might be expected to justify why 2k+2 is even.
 
  • Like
Likes sonadoramante
sonadoramante said:
Summary:: Prove that if a is an odd integer and b is an odd integer then a+b is even.

Theorem: If a is odd and b is odd then a+b is even.

Proof: Let a and b be positive odd integers of the form a = 2n+1 & b = 2m+1

a+b = 2n+1+2m+1
= 2n+2m+1+1
= 2n+2m+2
= 2(n+m)+2

Why not 2n + 2m + 2 = 2(n + m + 1)?

= Let k = n+m
= 2k+2
Therefore a+b is even.
 
  • Like
Likes docnet, Mark44 and sonadoramante
Aha! That makes more sense.
a+b = 2n+1+2m+1
= 2n+2m+2
= 2(n+m+1)
= 2k
Hence a+b is even. :)
 
sonadoramante said:
a+b = 2n+1+2m+1
= 2n+2m+1+1
= 2n+2m+2
= 2(n+m)+2
= Let k = n+m
= 2k+2
Therefore a+b is even.
The line "= Let k = n + m" shouldn't be there. The version in post #4 is what you want to say.
 
  • Like
Likes sonadoramante

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
View full post »

Similar threads

If p is even and q is odd, then ##\binom{p}{q}## is even
Replies
19
Views
3K
Prove that if a is even & b is any positive integer, then ab is even
Replies
9
Views
3K
If ## a ## is an odd integer, then ## a^{2}\equiv 1\pmod {8} ##?
Replies
2
Views
2K
Prove that ## x\equiv 1\pmod {2n} ##
Replies
2
Views
1K
Every integer greater than 5 is the sum of three primes?
Replies
4
Views
3K
Prove that the Goldbach conjecture that every even integer....
Replies
6
Views
2K
Quadratic equation with no rational roots
Replies
12
Views
3K
Show that 13 is the largest prime that can divide....
Replies
7
Views
2K
Prove that ## a^{({2}^{n})}\equiv 1\pmod {2^{n+2}} ## for any ##n##?
Replies
6
Views
2K
For ## n>1 ##, show that every prime divisor of ## n+1 ## is an odd integer
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top