Problem set with 5 proofs involving odd, even, parity, etc.

In summary, the conversation involves seeking help with a problem set and discussing proofs related to odd and even numbers, the irrationality of cube root 2, and a proof by induction. The moderator suggests providing a lemma and the person providing assistance suggests proving the irrationality of cube root 2 first.
  • #1
deme76
1
0
Homework Statement
1.Prove that if a and b are both odd, then a^2 b^2 is also odd.
2.Two integers are not the same parity if they are both even or both odd.
Prove that if x and y are of the same parity, then x+y is even.
3.Prove that if m-5 is odd, then (m-5)^(2 ) is odd.
4.Show that ∛2 is an irrational number.
5.Prove by induction that 1^2+ 2^2+⋯+ n^2= 1/6 (n)(n+1)(2n+1)
Relevant Equations
Show that ∛2 is an irrational number.
Assume ∛(2 ) rational
we can therefore say ∛2
= a⁄(b ) where a ,b are integers,and a and b are coprime
2= a^3/b^3
2b^3= a^3
hence,a is an even integers
we can say ,a=2n where m is an integer
〖2b〗^(3 )= (2m)^3
2b^3=8m^3
b^3= 〖4m〗^3
so b is also even.This complete the contradiction where we assumed
a and b were coprime.
Therefore, ∛2 is an irrational number
5.Prove by induction that 1^2+ 2^2+⋯+ n^2= 1/6 (n)(n+1)(2n+1)
help me in my problem set
qs
 
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  • #2
deme76 said:
help me in my problem set
You need to show some attempt at a solution. What have you tried?
 
  • #3
Moderator's note: Thread moved to precalculus math homework.
 
  • #4
PeterDonis said:
You need to show some attempt at a solution. What have you tried?
All his/her working is in the Relevant Equations section.
 
  • #5
deme76 said:
Homework Statement:: 1.Prove that if a and b are both odd, then a^2 b^2 is also odd.
2.Two integers are not the same parity if they are both even or both odd.
Prove that if x and y are of the same parity, then x+y is even.
3.Prove that if m-5 is odd, then (m-5)^(2 ) is odd.
4.Show that ∛2 is an irrational number.
5.Prove by induction that 1^2+ 2^2+⋯+ n^2= 1/6 (n)(n+1)(2n+1)
Relevant Equations:: Show that ∛2 is an irrational number.
Assume ∛(2 ) rational
we can therefore say ∛2
= a⁄(b ) where a ,b are integers,and a and b are coprime
2= a^3/b^3
2b^3= a^3
hence,a is an even integers
we can say ,a=2n where m is an integer
〖2b〗^(3 )= (2m)^3
2b^3=8m^3
b^3= 〖4m〗^3
so b is also even.This complete the contradiction where we assumed
a and b were coprime.
Therefore, ∛2 is an irrational number
5.Prove by induction that 1^2+ 2^2+⋯+ n^2= 1/6 (n)(n+1)(2n+1)

help me in my problem set
qs
I reckon that it would be better if you provide a lemma that if ##a^3## is even, then ##a## is even, and then go for proving that cube root 2 is irrational. The reason for that is, it is standard to assume ##a## to be even when ##a^2## is given to be even, but the case of cube is not standard, so, we should prove it first.
 

FAQ: Problem set with 5 proofs involving odd, even, parity, etc.

What is the definition of an odd number?

An odd number is any integer that cannot be divided evenly by 2, resulting in a remainder of 1. In other words, an odd number is not a multiple of 2.

How do you prove that a number is even?

To prove that a number is even, you can divide it by 2 and if the remainder is 0, then the number is even. Another way to prove that a number is even is to show that it can be written in the form 2n, where n is an integer.

What is the definition of parity?

Parity refers to whether a number is even or odd. It is determined by the remainder when the number is divided by 2. If the remainder is 0, the number is even and if the remainder is 1, the number is odd.

How can you prove that the sum of two odd numbers is even?

To prove that the sum of two odd numbers is even, you can use the fact that an odd number can be written as 2n+1, where n is an integer. So, the sum of two odd numbers would be (2n+1) + (2m+1) = 2(n+m+1), which is in the form of 2n, making it an even number.

Can you use proof by contradiction to prove that a number is odd?

Yes, you can use proof by contradiction to prove that a number is odd. Assume that the number is even, then it can be written as 2n, where n is an integer. But this would mean that the number is divisible by 2, contradicting the definition of an odd number. Therefore, the number must be odd.

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