steps of proving irrationality
Answers
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.
Answer:
Let’s look at Euclid’s proof that the square root of two is irrational.
Let us assume that the root of two is rational and represent it as such:
2–√=pq
$p$ and q are assumed to be co-prime integers, i.e., the fraction is fully reduced.
Solve for p and square both sides:
p2=2q2
We see that p must be even, since its square is even, making q odd since they are co-prime. We can then let p=2m and q=2n+1 to represent an even and odd number respectively.
(2m)2=2(2n+1)2
Expansion of the squares yields:
4m2=2(4n2+4n+1)
Divide through by 2 and partially factor the right side:
2(m2)=2(2(n2+n))+1
Let u=m2 and v=2(n2+n) and substitute:
2u=2v+1
Here we absurdly have an even natural number being equal to an odd one. This impossibility means without a doubt that the assumption we made regarding the rationality of the square root of two is false. We can also look at it this way by dividing through by 2:
u=v+12
This of course is also impossible for natural numbers, to be one half unit apart. This is what's known as proof by reductio ad absurdium. You begin with an assumption and then reduce the statement to an absurdity, thereby proving the assumption to be false.
This is essentially the proof given by Euclid in his The Elements.