Download For all x there exists ay such that for all z, if z>y then z>x+y. If z>y

Questions on Exercise 2.6:
For all x there exists a y such that for all z, if z > y then z > x+y.
If z > y then z > 5 + y. False.
If z > y then z > 0 + y. True.
Or x could be anything less than zero.
Even for a specific y, (for all z, z > y implies z > x + y) means
x ≤ 0.
So effectively the statement is saying “for all x, x ≤ 0”. Which
isn’t true.
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For all x and for all y there exists z such that z > y implies
z > x + y.
This is true.
The difference here is that the z is existentially quantified instead
of universally, so you get to choose z. So you can choose z so
that z ≤ y, which makes the implication true. Furthermore you
can do this for any x and y.
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Comments on homework An equation is not a statement, it’s
part of a statement. The equation
x2 + x − 2 = 0
could fit into a number of different statements. For example:
The solutions to
x2 + x − 2 = 0
are −2 and 1.
The number 5 is not a solution to the equation
x2 + x − 2 = 0.
In the first statement, x is universally quantified: it says “for all
x such that x2 + x − 2, x = −2 or x = 1. In the second, x is
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not really a variable, it’s just another name for 5. Moral: Have
to say what x is or quantify it some way or another. You can’t
just have a free x floating around. In the case of the homework
problem, for example, start your proof with “Let x be a solution
to . . . .”
There’s a difference between an if-then statement and it’s
converse.
“If x = 2 or x = −1, then x2 + x − 2 = 0”
is not the same as
“If x is a solution then x = 2 or x = −1.”
and
“If a = 0 or b = 0 then ab = 0. ”
is not the same as
“If ab = 0 then a = 0 or b = 0.”
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The mechanical procedures for solving an equation can serve
many different lines of mathematical reasoning. Your goal is to
bring out those hidden lines.
Example:
For all real numbers y 6= 1, there exists a real number x such
that (x + 1)/x = y.
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-10
10
-10
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Proof Let y be a real number not equal to 1, and set x =
1/(y − 1). Then
1
1 + (y − 1)
x+1
y−1 + 1
=
=
= y.
1
x
1
y−1
This is an example of a proof which was constructed backwards.
First we solved the equation for x, then stated the solution at
the beginning of the proof.
If you want to prove that a number satisfying an equation exists,
the most direct way to do so is to exhibit the number up front.
The work you did to solve the equation is not a necessary part
of the proof. Once you have the number in hand, all you have
to do is put it into the equation to show that it’s a solution.
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Write converse and contrapositive of
If it’s an apple, it’s red.
Converse If it’s red then it’s an apple. Not an equivalent statement.
Contrapositive If it’s not red then it’s not an apple.
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Example Irrationality of
Direct statement: If x =
√
2.
√
2, then x is irrational.
Contrapositive: If x is rational, then x2 6= 2.
The second formulation is easier because we have something to
work with in starting the proof: “Suppose x is rational. Then
x = a/b where a and b are integers (b 6= 0). etc.”
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Find contrapositive and converse of
• If xz = yz and z 6= 0, then x = y.
Contrapositive: If x 6= y, then xz 6= yz or z = 0.
Converse: If x = y, then xz = yz and z 6= 0.
• If xy = 0, then x = 0 or y = 0.
• If x 6= y then x2 6= y 2.
Which is true and which false?
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Proof by contrapositive:
If 2n + 1 is a prime number then n is even.
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