QUESTION 1
a) Use a proof by cases to show that 27 is not the square of a positive integer. b)
What is wrong with the following “proof” that -3 = 3, using backward reasoning?
Assume that -3 = 3. Squaring both sides yields (-3)2 = 32 or 9 = 9. Therefore - 3 =
3.
a) To prove that 27 is not the square of a positive integer, we can consider the
possible cases:
Case 1: The square root of 27 is a positive integer. If the square root of 27 is a
positive integer, then there exists an integer, let's say "x," such that x^2 = 27. However,
taking the square root of both sides, we have √(x^2) = √27, which simplifies to |x| =
√27. Since x is a positive integer, |x| must also be a positive integer. But the square
root of 27 (√27) is an irrational number (approximately 5.196), and therefore not an
integer. This contradicts our assumption. Hence, 27 cannot be the square of a
positive integer in this case.
Case 2: The square root of 27 is a negative integer. If the square root of 27 is a
negative integer, then there exists an integer, let's say "y," such that y^2 = 27.
However, taking the square root of both sides, we have √(y^2) = √27, which simplifies
to |y| = √27. Since y is a negative integer, |y| will be a positive integer. However, the
square root of 27 (√27) is an irrational number (approximately 5.196), and therefore
not an integer. This contradicts our assumption. Thus, 27 cannot be the square of a
positive integer in this case either.
Since we have exhausted all possible cases and arrived at a contradiction in each
case, we can conclude that 27 is not the square of a positive integer.
b) The "proof" that -3 = 3 using backward reasoning is flawed because it assumes
the statement (-3)^2 = 3^2 implies -3 = 3. While it is true that if two numbers have the
same square, then they must be either both positive or both negative, it does not
, necessarily mean that the numbers themselves are equal. In this case, squaring both
sides of -3 = 3 yields (-3)^2 = 3^2, which simplifies to 9 = 9, and this is indeed true.
However, it does not imply that -3 and 3 are equal. The flaw lies in assuming that if
the squares of two numbers are equal, then the numbers themselves must be equal.
This assumption is incorrect, and therefore the "proof" is invalid.
QUESTION 2
a) Prove: if m and n are even integers, then mn is a multiple of 4. b) Prove the
following theorem: n is even if and only if n2 is even. c) Prove that the following is
true for all positive integers n: n is even if and only if 3n2 + 8 is even.
a) Proof: Suppose m and n are even integers. By definition, an even integer can be
expressed as 2k, where k is an integer. Therefore, we can write m = 2a and n = 2b for
some integers a and b.
Now, let's consider the product mn: mn = (2a)(2b) = 4ab
Since a and b are integers, their product 4ab is also an integer. Moreover, we can
express 4ab as 4(ab), which is a multiple of 4.
Hence, we have shown that if m and n are even integers, then mn is a multiple of 4.
b) Proof: (i) If n is even, then n^2 is even. Suppose n is an even integer. By definition,
an even integer can be expressed as 2k, where k is an integer. Therefore, we can
write n = 2k for some integer k.
Now, let's consider the square of n: n^2 = (2k)^2 = 4k^2
Since k is an integer, k^2 is also an integer. Moreover, we can express 4k^2 as 4(k^2),
which is a multiple of 4.
Hence, if n is even, then n^2 is even.
a) Use a proof by cases to show that 27 is not the square of a positive integer. b)
What is wrong with the following “proof” that -3 = 3, using backward reasoning?
Assume that -3 = 3. Squaring both sides yields (-3)2 = 32 or 9 = 9. Therefore - 3 =
3.
a) To prove that 27 is not the square of a positive integer, we can consider the
possible cases:
Case 1: The square root of 27 is a positive integer. If the square root of 27 is a
positive integer, then there exists an integer, let's say "x," such that x^2 = 27. However,
taking the square root of both sides, we have √(x^2) = √27, which simplifies to |x| =
√27. Since x is a positive integer, |x| must also be a positive integer. But the square
root of 27 (√27) is an irrational number (approximately 5.196), and therefore not an
integer. This contradicts our assumption. Hence, 27 cannot be the square of a
positive integer in this case.
Case 2: The square root of 27 is a negative integer. If the square root of 27 is a
negative integer, then there exists an integer, let's say "y," such that y^2 = 27.
However, taking the square root of both sides, we have √(y^2) = √27, which simplifies
to |y| = √27. Since y is a negative integer, |y| will be a positive integer. However, the
square root of 27 (√27) is an irrational number (approximately 5.196), and therefore
not an integer. This contradicts our assumption. Thus, 27 cannot be the square of a
positive integer in this case either.
Since we have exhausted all possible cases and arrived at a contradiction in each
case, we can conclude that 27 is not the square of a positive integer.
b) The "proof" that -3 = 3 using backward reasoning is flawed because it assumes
the statement (-3)^2 = 3^2 implies -3 = 3. While it is true that if two numbers have the
same square, then they must be either both positive or both negative, it does not
, necessarily mean that the numbers themselves are equal. In this case, squaring both
sides of -3 = 3 yields (-3)^2 = 3^2, which simplifies to 9 = 9, and this is indeed true.
However, it does not imply that -3 and 3 are equal. The flaw lies in assuming that if
the squares of two numbers are equal, then the numbers themselves must be equal.
This assumption is incorrect, and therefore the "proof" is invalid.
QUESTION 2
a) Prove: if m and n are even integers, then mn is a multiple of 4. b) Prove the
following theorem: n is even if and only if n2 is even. c) Prove that the following is
true for all positive integers n: n is even if and only if 3n2 + 8 is even.
a) Proof: Suppose m and n are even integers. By definition, an even integer can be
expressed as 2k, where k is an integer. Therefore, we can write m = 2a and n = 2b for
some integers a and b.
Now, let's consider the product mn: mn = (2a)(2b) = 4ab
Since a and b are integers, their product 4ab is also an integer. Moreover, we can
express 4ab as 4(ab), which is a multiple of 4.
Hence, we have shown that if m and n are even integers, then mn is a multiple of 4.
b) Proof: (i) If n is even, then n^2 is even. Suppose n is an even integer. By definition,
an even integer can be expressed as 2k, where k is an integer. Therefore, we can
write n = 2k for some integer k.
Now, let's consider the square of n: n^2 = (2k)^2 = 4k^2
Since k is an integer, k^2 is also an integer. Moreover, we can express 4k^2 as 4(k^2),
which is a multiple of 4.
Hence, if n is even, then n^2 is even.
