Explain steps of proving irrationality
Answers
proving √2 is irrational -
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.
From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction......
If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:
2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!
WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.
Answer:
Assume that a is rational, b is irrational, and a + b is rational. Since a and a + b are rational, we can write them as fractions.
Let a = c/d and a + b = m/n
Plugging a = c/d into a + b = m/n gives the following:
c/d + b = m/n
Now, let's subtract c/d from both sides of the equation.
b = m/n - c/d, or
b = m/n + (-c/d)
Since the rational numbers are closed under addition, b = m/n + (-c/d) is a rational number. However, the assumptions said that b is irrational, and b cannot be both rational and irrational. This is our contradiction, so it must be the case that the sum of a rational and an irrational number is irrational.
And that's our proof!
There's only one more sum to consider, and that is the sum of two irrational numbers.