Final answer:
To show that the curve x²+y²−3=0 has no rational points, we can use contradiction. Assume that there exists a rational point (a,b) on the curve. Substituting these values into the equation x²+y²−3=0, we can find a contradiction by showing that the left and right sides of the equation cannot both be integers squared.
Step-by-step explanation:
To show that the curve x²+y²−3=0 has no rational points, we can use contradiction. Assume that there exists a rational point (a,b) on the curve. This means that a and b can be expressed as fractions in the form a=p/q and b=r/s, where p, q, r, and s are integers and s and q are not zero.
Substituting these values into the equation x²+y²−3=0, we get (p/q)²+(r/s)²−3=0. Multiplying both sides by q²s², we have p²s²+r²q²−3q²s²=0.
Simplifying the equation, we have p²s²−3q²s²=-r²q². Since p, q, r, and s are all integers, the left side of the equation is an integer. But the right side of the equation is the square of an integer (-r²q²), which means it is also an integer. This implies that p²s²−3q²s²=-r²q² is an equation of an integer equal to the square of another integer. However, there is no such case, which contradicts our assumption. Therefore, the curve x²+y²−3=0 has no rational points.