Spring Semester 2016, CS160 1. (25 points) Direct Proof: Use a direct proof to s

Question

Spring Semester 2016, CS160

1. (25 points) Direct Proof: Use a direct proof to show that 4a + 5b is odd when a is an even integer and b is an odd integer. Note: Use as many steps as necessary.

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Contrapositive Proof: Use a contrapositive proof to show that if a and b are integers, and 5ab is even, then a is even or b is even. Note: Use as many steps as necessary.

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Contradiction Proof: Use a contradiction proof to show that if a and b are integers, and 3ab is even, then a is even or b is even. Note: Use as many steps as necessary.

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Explanation / Answer

1)
4a + 5b is odd when a is even integer and b is odd integer
proof:
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consider 4a .. what ever the integer it gives always even integer
for 5b ... if integer is odd, it given odd number.
So if we add even + odd number .. it gives finally odd number.
So proof completed.

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2)
Let us proof this by contradiction.
so Assume a and b are integers and 5ab is even , then a is not even or b is not even

if 5ab is even ... then ab product should be even ...
for even the chances are:
even x even = even
even x odd = even
odd x even = even

So this shows either one of the number (a or b) must even.
So contradiction occurs as per our assumption.
So, if 5ab is even then a or b must be even. proof completed.

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3)
Let us proof this by contradiction.
so Assume a and b are integers and 3ab is even , then a is not even or b is not even

if 5ab is even ... then ab product should be even ...
for even the chances are:
even x even = even
even x odd = even
odd x even = even

So this shows either one of the number (a or b) must even.
So contradiction occurs as per our assumption.
So, if 3ab is even then a or b must be even. proof completed.

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