prove that there are no integars p and q such that:
(p/q)squared =2
To prove that there are no integers p and q such that (p/q)^2 = 2, we can use proof by contradiction.
First, assume that there exist integers p and q such that (p/q)^2 = 2. This implies that p^2/q^2 = 2, which can be rearranged to p^2 = 2 * q^2.
Now, we can analyze the equation p^2 = 2 * q^2. For p^2 to be an even number (divisible by 2), p must also be even. So, we can write p = 2 * k, where k is an integer.
Substituting p = 2 * k into the equation p^2 = 2 * q^2, we get (2 * k)^2 = 2 * q^2, which simplifies to 4 * k^2 = 2 * q^2. Dividing both sides by 2, we have 2 * k^2 = q^2.
Now, let's analyze the equation q^2 = 2 * k^2. Similar to before, q^2 must also be even, which means q must be even. We can write q = 2 * m, where m is an integer.
Substituting q = 2 * m into the equation q^2 = 2 * k^2, we get (2 * m)^2 = 2 * k^2, which simplifies to 4 * m^2 = 2 * k^2.
Dividing both sides by 2, we have 2 * m^2 = k^2.
Now, we have shown that if p^2 = 2 * q^2, then k^2 = 2 * m^2.
However, this contradicts the assumption that p and q are integers since we have shown that both p and q must be even. If both p and q are even, they have a common factor of 2, making their ratio (p/q) not an integer.
Therefore, our assumption that there exist integers p and q satisfying (p/q)^2 = 2 is false. Hence, we have proven that there are no integers p and q that satisfy the given equation.