efine ψ = √5−1 2 . for positive real integers a, b with a ≤ b, check whether b/(a + b) is closer to ψ than a/b. explore how large you can choose a, b before computation time reaches a minute. in matlab

main.m
% Define ψ = (√5 - 1) / 2
psi = (sqrt(5) - 1) / 2;

% Initialize variables
a = 1;  % Starting value for a
b = 1;  % Starting value for b
max_a = 1;  % Maximum value of a to be tested
max_b = 1;  % Maximum value of b to be tested
min_difference = 1;  % Initialize the minimum difference

tic;  % Start timer

while toc < 60  % Run the loop until computation time reaches 1 minute
    % Calculate the differences for both ratios
    diff_ratio1 = abs(psi - a / b);
    diff_ratio2 = abs(psi - b / (a + b));
    
    % Check if diff_ratio2 is smaller than diff_ratio1
    if diff_ratio2 < diff_ratio1
        fprintf('b/(a + b) is closer to psi than a/b for a = %d and b = %d\n', a, b);
    end
    
    % Update the minimum difference if necessary
    min_difference = min(min_difference, diff_ratio2);
    
    % Increment a and b
    a = a + 1;
    b = b + 1;
    
    % Check if a or b has reached the maximum value
    if a > max_a || b > max_b
        break;
    end
end

fprintf('The minimum difference is %f\n', min_difference);
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