Prove that the sum of 3/4 and √10 is irrational. Show your work, and in 1–2 sentences, explain why the sum is an irrational number.(2 points)
Let's assume that the sum of 3/4 and √10 is a rational number, which means it can be written as a fraction a/b where a and b are integers and b is not equal to 0. Then, 3/4 + √10 = a/b. We can rearrange this equation to get √10 = (a/b) - 3/4 = (4a - 3b)/4b. This implies that √10 is rational, but since √10 is actually irrational, our assumption is false. Therefore, the sum of 3/4 and √10 is an irrational number.
To prove that the sum of 3/4 and √10 is irrational, we will assume the opposite, then derive a contradiction.
Let's suppose that the sum of 3/4 and √10 is rational. This means it can be expressed as a fraction p/q, where p and q are integers and q is not equal to zero.
Therefore, we have (3/4) + (√10) = p/q.
Rearranging the equation, we get √10 = (p/q) - (3/4).
Squaring both sides of the equation, we have 10 = (p/q)^2 - (3/4)^2.
Simplifying further, we get 10 = (p^2)/(q^2) - 9/16.
Multiplying both sides by 16q^2, we have 160q^2 = 16p^2 - 9q^2.
This implies that 9q^2 = 16p^2 - 160q^2.
Rearranging the equation, we get 16p^2 = 169q^2.
This shows that 169q^2 is divisible by 16 and hence, q^2 is divisible by 16.
Since q is an integer, q^2 must be divisible by 16, and thus, q must be divisible by 4.
Let's consider q = 4r, where r is an integer.
Substituting this value back into the equation, we have 169(4r^2) = 16p^2.
Dividing both sides by 16, we get 169r^2 = p^2.
This implies that p^2 is divisible by 169, and thus, p must be divisible by 13.
As both p and q are divisible by 13 and 4, respectively, they have a common factor, which contradicts our initial assumption that p/q is a fraction in simplest form.
Therefore, our assumption that the sum of 3/4 and √10 is rational is false, and hence, the sum is an irrational number.