Proof by Contradiction

[Guest]

I must prove these by contradiction though I know it would be easier to do it another way. And, I must follow formal steps of a proof as this is a proof writing class.

1) If m is odd and n is odd, then 4|\(does not divide) m^2+n^2
my work:
Assume m and n are odd
Assume 4|m^2+n^2
Since m is odd, by definition, m=2k+1
Since n is odd, by definition, n=2t+1
Since 4|m^2+n^2, by definition, m^2+n^2=4j
k,t, and j are all integers. this is where I get stuck.

2)If n is odd, then 7n-5 is even
Assume n is odd
Assume 7n-5 is odd
Since n is odd by def. n=2k+1
since 7n-5 is odd, by def. 7n-5=2t+1
by substitution, 7(2k+1)-5=2t+1
14k+7-5=2t+1
14k+2=2t+1
1=2(t-7k)
Let j=t-7k j is an integer
So, 1=2j
by definition, 1 is even (obviously not true)
But, 1= this is where I get stuck

3)If x is an element of real numbers and x does not equal 0 and x+1/x<2 then x<0
Assume x is an element of real numbers
Assume x does not equal 0
Assume x+1/x<2
Assume x>0
I have no clue where to begin on this proof
HELP! Thanks.

 

[pka]

For #1, look at the following;
\(\displaystyle \begin{array}{rcl}
\left( {2k + 1} \right)^2 + \left( {2j + 1} \right)^2 & = & \left( {4k^2 + 4k + 1} \right) + \left( {4j^2 + 4j + 1} \right) \\
& = & 4\left( {k^2 + j^2 } \right) + 4\left( {k + j} \right) + 2 \\
\end{array}\)

For #3, let x=1.

 

[Guest]

if x=0 in #3, then the statement x+1/x<2 would be false but that is not where the contradiction should be.

 

[Unco]

I must prove these by contradiction though I know it would be easier to do it another way. And, I must follow formal steps of a proof as this is a proof writing class.

1) If m is odd and n is odd, then 4|\(does not divide) m^2+n^2
my work:
Assume m and n are odd
Assume 4|m^2+n^2
Since m is odd, by definition, m=2k+1
Since n is odd, by definition, n=2t+1
Since 4|m^2+n^2, by definition, m^2+n^2=4j
k,t, and j are all integers. this is where I get stuck.

The next logical step would be to substitute your expressions for m and n into m^2 + n^2 = 4j. You can show from there that j cannot be an integer, a contradiction.

2)If n is odd, then 7n-5 is even
Assume n is odd
Assume 7n-5 is odd
Since n is odd by def. n=2k+1
since 7n-5 is odd, by def. 7n-5=2t+1
by substitution, 7(2k+1)-5=2t+1
14k+7-5=2t+1
14k+2=2t+1
1=2(t-7k)
Let j=t-7k j is an integer
So, 1=2j
j is an integer so j = 1/2 is a contradiction.
by definition, 1 is even (obviously not true)
But, 1= this is where I get stuck

3)If x is an element of real numbers and x does not equal 0 and x+1/x<2 then x<0
Assume x is an element of real numbers
Assume x does not equal 0
Assume x+1/x<2
Assume x>0
I have no clue where to begin on this proof
Although a counterexample is the quickest, you could also derive a contradiction by multiplying both sides of the inequality by x, assuming that x>0 so the direction of the inequality does not change. This will give rise to a contradiction as a real number cannot be squared to give a negative number.
HELP! Thanks.

 

[Guest]

The next logical step would be to substitute your expressions for m and n into m^2 + n^2 = 4j. You can show from there that j cannot be an integer, a contradiction.

I understand the whole substitution and thus I get (2k+1)^2 + (2t+1)^2=4j
What I don't understand is how to contradict If m is odd and n is odd, then 4|\(does not divide) m^2+n^2

My contradiction is not that j is not an integer, but that statement above.

The same with number 2. I am stating that j is an integer based on the fact that it is the sum of the integers above. It must be an integer because it is closed under addition and multiplication.
So in #2 my contradiction is not that j is not an integer but it is supposed to contradict the statement If n is odd, then 7n-5 is even

Thanks!

 

[daon]

The next logical step would be to substitute your expressions for m and n into m^2 + n^2 = 4j. You can show from there that j cannot be an integer, a contradiction.

I understand the whole substitution and thus I get (2k+1)^2 + (2t+1)^2=4j
What I don't understand is how to contradict If m is odd and n is odd, then 4|\(does not divide) m^2+n^2

My contradiction is not that j is not an integer, but that statement above.

Your statement is true IF AND ONLY IF j is an integer. Contradict one and you also contradict the other. When you use definitions to prove your statement, they are all IFF statements, so any derived contradiction will do.