Consider the following statement and "proof". 

For a real number x, if x>0 then (x^2+2)/(4x) >0 . 

Proof:

1. Suppose x>0 but (x^2+2)/(4x) \leq 0. 
2. Then we may multiply both sides of (x^2+2)/(4x) by x without changing the inequality, since x>0.
3. Hence (x^2+2)/4 \leq 0.
4. Then (x^2+2) \leq 0 . 
5. So x^2 \leq -2 . 
6. But x>0 implies x^2>0. 
7. This contradiction proves our original assumption was false, so the statement is true. 

Which of the following are true?

Choose the converse of the following statement:

"If p is a prime number then 2^p+1 is also a prime number".

Let n be a positive integer. Consider the statement:

"If 7n is an odd integer, then n is an odd integer."

Choose from below the contrapositive of this statement.

Consider the statement:

"There exists an integer x, such that 4x-x^2>3."

Prove the statement by finding the integer that allows it to be true.

Consider the following statement and its "proof". 

If m is an even integer and n is an odd integer, then 3m + 5n is odd.

Proof. 

1. Let m be an even integer and n an odd integer. 
2. Then there is an integer k such that m = 2k and n = 2k +1.  
3. Therefore 3m + 5n = 3(2k) + 5(2k + 1) = 16k+5 = 2(8k+2)+1
4. Since 8k + 2 is an integer, 3m + 5n is odd.

Which of the following is correct?

Consider the statement “If 3 divides (x^2 + 1), then 3 does not divide x.” 

1. Assume 3 divides x. Then x = 3k for some integer k.  Hence 

2. Assume 3 divides (x^2 + 1). So we can write x^2 + 1 = 3k for some integer k.

3. Assume 3 is not divisible by x. So we can write x = 3k + 1 or 3k + 2 for some integer k.

Which of the three proofs is correct?

Consider the following proof.

Proof: Assume that there is an odd integer n which can be expressed as the sum of three even integers x, y and z. Then x = 2a, y = 2b, and z = 2c where a, b, c are integers. Therefore

This is a proof of what statement?

Consider the statement:

"For any integers x and y, if xy+x is odd, then x is odd." 

The following is a "proof": 

Proof:

1. Let x and y be integers.
2. If x is odd, then there is an integer p such that x=2p+1.
3. Then xy+x = (2p+1)y+(2p+1) = 2(py+p) + y+1 . 
4. But if y is odd, then 2(py+p)+y+1 is even. 
5. Thus the statement is false. 

Which of the following is true?

Which is the negation of the following statement:

For all real numbers x, x^2+x > 0.