From Step 3, you have . So is a multiple of . This means is a multiple of . This can be proven using the contrapositive.
So we can let and therefore
.
Since is a multiple of , so is .
There is your common factor and contradiction.
Hi
I have this worked example to prove is an irrational number via contradiction. I will put it down here now. But my question is how do I do this with as I get stuck half way through this when we come to talking about even numbers.
Step 1: Suppose that is rational.
Step 2: This means that its possible to find where a and b have only a common factor of 1 such that:
, b not equal to 0.
Squaring both sides:
Step 3:
This means that must be even and thus must be even.
i.e.
Step 4:
This means that b^2 and b must be even.
Step 5: So both a and b are even. This means that a and b have a common factor or 2. This contradicts the original hypothesis and so the hypothesis must be false. This means that is not able to be expressed in the form a/b and is therefore an irrational number.
The end.
So that is the example proof. Here is what I have done so far with the example.
Step 1: Suppose that is rational.
Step 2: This means that its possible to find where a and b have only a common factor of 1 such that:
, b not equal to 0.
Squaring both sides:
Step 3:
This condition is only met when a is odd and b is odd. I just did this with a little trial and error. So a is odd. Thus a = 2k + 1
Step 4:
......
Ok this is where I come unstuck. Does anyone have any suggestions?
David.
Here is a proof that a multiple of 3 implies is a multiple of 3 (by contrapositive).
Suppose a is not a multiple of 3. Then or for some integer . If , then . is an integer because the integers are closed under addition and multiplication. So is not divisible by 3. The second case is similar.