We did this by contradiction, assuming that such a fraction existed, noting that without loss of generality that that fraction could be written in lowest terms so that a and b are co-prime (they share no factors). In particular, this means that a and b CANNOT both be even. This gives us exactly 3 possibilities for the parity of a and b. If we rewrite a/b=√2 as a=b√2, and if we look carefully, we can arrive at a contradiction in each of the three cases.
For this problem, I’m going to assume that you’ve done that, and know for a rock-solid fact that 2 is irrational. Given that fact:a. Show that 7+√2 is an irrational number.
b. Show that 7+√2 is an algebraic number. To do b., some additional background may be helpful.
Background for b.
We have seen rational numbers and irrational numbers. Rational numbers come from arithmetic of whole numbers, and our first source of irrational numbers is geometry. All the numbers that we can make with a compass and straightedge are called constructable numbers. In more modern times, we have another source of numbers: polynomial equations with rational numbers for their coefficients. Numbers who are solutions to these equations are called algebraic numbers.
Here are a couple of examples:
1) 2 is an algebraic number because it's a solution to the equation x2-2=0. If you plug 2 in for x, the equation is true.
2) 5 is the solution to x-5=0. So it too is an algebraic number. Duh.
3) Even though 2 is algebraic, it's not because x-2=0. To show that a number is algebraic, you can't use irrational numbers in the equation.
So, let's first start with 8a, and let's show that 7 + √2 is an irrational number.
First, Ben reminded me to assume that 7 + √2 is equal to a/b.
So:
7 + √2 = a / b
7b + √2b = a
√2b = a - 7b Now, let's think about it. Is there anyway for an irrational number multiplied to an integer to create an integer? No! So this means that any result on the left side of the equation has to be an irrational number. Now, let's look at the right side. What does 7 times any rational number make? Another rational number. So, is there anyway for a rational number minus another rational number to create an irrational number? No! Because any rational subtracted from another rational number will create a rational number.
This means that the left side is an irrational number and the right side is an rational number. There is no way for number to be both rational and irrational, so 7 +√2 is an irrational number.
Then for 8b. How do we know whether or not 7 +√2 is an algebraic number? Well, we'd first have to define what an algebraic number is. According to the backgound, algebraic numbers are numbers who are solutions to polynomial equations with rational numbers for their coefficients. So, there are two requirements, algebraic numbers must be a solution to an algebraic equation, and this equation must only have rational numbers like fractions and whole numbers.
So, to show that this is possible, let's work backwards and assume there is a problem in which the solution is:
x = 7 +√2
x - 7 = √2
(x-7)2= 2
(x-7)2 - 2 = 0
x2 - 14x + 49 -2 = 0
x2 - 14x + 47 = 0
I was able to work backwards and produce an algebraic equation that has only rational numbers and, because, I worked backwards, to make sure that 7 + √2 is the solution to the equation then 7 + √2 has to be an algebraic number because I've found an algebraic equation that has only rational numbers and in which 7 + √2 is the solution.
Bonus: Prove that √3 is irrational
To prove that √3 is irrational, I first have to assume that it is rational.
If that is the case then there will be a fraction such that two non-common factors a and b divided by each other that is equal to the √3:
√3 = a/b
(√3 = a/b)2
3 = a2 / b2
3b2 = a2
So, let's consider all the possibilities,
If a is odd and b is even then:
3(even #)2 = (odd #)2
3 x even # = odd #
even # = odd #
This is not possible, a number can't both be odd and even at the same time, so this is not possible.
What if a is even and b is odd, then:
3(odd #)2 = (even #)2
3 x odd # = even #
odd # = even #
The same problem happens again and, like before, a number can't both be odd and even at the same time, so this is not possible.
Alright, so what if both a and b were even, this there will be no problem then right?
3 (even #) = (even #)
3 x even # = even #
even # = even #
However, if two numbers are even one can always divide them by 2 and so our initial assumption that a/b have no common factor is wrong, because if they do, they can further be simplified. I can then imply that a and b cannot be both even.
Now, what if both a and b were odd?
Well, if they both were odd then I would be able to rewrite them like this:
a = 2s + 1
b = 2t + 1
The s and t are required to be integers, because this is the only way to ensure that a and b are integers.
Now, let's plug these in for a and b:
3(2t+1)2 = (2s+1)2
3(4t2 + 4t + 1) = (4s2 + 4s + 1)
12t2 + 12t + 3 = 4s2 + 4s + 1
(12t2 + 12t + 2 = 4s2 + 4s) x 1/2
6t2 + 6t + 1 = 2s2 + 2s
If I look into this, any integer that I plug in for t on the left will create an odd number, and because there's not a +1 on the right side of the equation, any integer I plug in for s will create an even number. Like before, there's no way for a number to be both odd and even at the same time so a and b can't both be odd.
So, there are no combinations of a and b (made of non-common factors) such that a/b is equal to √3. If there are no such cases, then that means the √3 is irrational.
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