Conditional Statements

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To show me homework solutions during meetings, the best thing to do is get it onto your computer and screen-share it. For the first assignment, writing it by hand on paper, taking a photo of the paper, and uploading to your computer should work. Later on though, I’ll need typewritten solutions, which you can produce with Word, or something called LaTeX that I’ll introduce you to soon.

Conditional Statements

From “Conditional Statements” in section 1.1 and the conditionals discussion.

Truth and Falsehood of Conditionals

Are the following conditional statements true or false? Why?

1. If x is a real number, then x+1 is a real number.
   • This looks like an “obvious” conditional, where the hypothesis somehow causes or leads to the conclusion, and therefore it seems true.
   • It is indeed true, but truth and falsehood of conditionals is solely based on whether the hypothesis and conclusion are true and false in a pattern that makes the whole conditional true according to the “P → Q” truth table; it has nothing to with any humanly meaningful relationship between the hypothesis and conclusion.
   • In this case we have to consider 2 rows in the truth table. In the first, x is real, and so the hypothesis is true. The conclusion is also true, and so the whole conditional is true. Alternatively, maybe x isn’t real, in which case the hypothesis is false, and so the whole conditional is immediately true. In both cases the statement as a whole is true though.
2. If 1.5 is a real number, then 2.5 is a real number.
   • True, because the hypothesis and conclusion are both true.
3. If 1.5 is a real number, then 2.5 is an integer.
   • False the hypothesis is true but the conclusion is false.
4. If x is a real number, then 3 is an integer.
   • True, as analyzed in the first example.
   • It doesn’t matter that there’s no connection between x being real and 3 being an integer.
5. If 1.5 is an integer, then pigs fly.
   • True, because the hypothesis and conclusion are both false.

Suppose I tell you that “if π² is rational, then it’s raining gold coins” is a true statement. You look out the window and see that it is not raining gold coins. What can you conclude about π²?

That π² must not be rational, because the only way a true conditional can have a false conclusion is if the hypothesis is also false.

What if when you look out the window it is raining gold coins?

This doesn’t tell you anything, because a conditional with a true conclusion would be true regardless of whether the hypothesis were true or false.

Truth Rules and Proof

What do you need to do to show that a conditional is false? Why? (For example, to prove that “if x and y are integers such that x < y, then there is another integer, c, such that x < c < y” is not true.)

Show that it’s possible for the hypothesis to be true but the conclusion to be false, because that’s the only situation under which a conditional is false.

In this case we could make the hypothesis true by letting x and y be 4 and 5, respectively. But then the conclusion is false, because there’s no integer in between 4 and 5.

What do you need to do to show that a conditional is true? Why? (For example, to prove that it is true that “if x and y are real numbers such that x < y, then there is another real number, c, such that x < c < y.”)

At first, it seems you need to check each of the other 3 lines in the truth table, i.e., make sure it’s possible to have the hypothesis be false and the conclusion true, hypothesis false and conclusion false, etc. But thinking a little more about this, we really don’t have to worry about the cases where the hypothesis is false, because the conditional is automatically true then. The only case there’s any work for is when the hypothesis is true, when we need to show that the conclusion necessarily has to be true too. This thinking justifies the usual way of proving a conditional, namely by assuming that everything in the hypothesis actually is true, and showing that the conclusion must then also be true.

For example, we could use this idea to fairly informally prove that if x and y are real numbers such that x < y, then there is another real number, c, such that x < c < y.

A good strategy might be to start by assuming that x and y really are real numbers, and that x really is less than y, and seeing if we could find or calculate a c in between x and y.

Letting c be the average of x and y, i.e., c = (x+y)/2, should work, because…

From our past understanding of averages, an average is greater than the smallest number included in the average, as long as the numbers aren’t all equal. Since x < y, x and y aren’t equal, so x < (x+y)/2

Similarly, an average is less than the largest number included in it (as long as the numbers aren’t all equal), so (x+y)/2 < y.

Next

“Direct” proofs, i.e., proofs that show how the conclusion to some conditional follows directly through some chain of deductions from the hypothesis.

Read “Closure Properties of Number Systems” in section 1.1 of the textbook, and Beginning Activities 1 and 2, “Properties of Number Systems,” and “Constructing a Proof of a Conditional Statement” in section 1.2.

Participate in this direct proofs discussion to begin using the ideas from the reading.

For people whose cohort doesn’t get to come to class (e.g., Cohort A for Wednesday, Cohort B for today) you still need to

• Do the reading
• Participate in the Canvas discussion
• Review notes from the class meeting(s) you can’t come to
• Ask any questions need need to when you next come to class!

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