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   "source": [
    "# Tutorial A2 – Solutions"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "__1__ Consider the equation $y = \\frac{3a + 2b}{b^2}$, with $a = 1$ and $b = 2$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " - Calculate $y$ using Python."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.75\n"
     ]
    }
   ],
   "source": [
    "a = 1\n",
    "b = 2\n",
    "y = (3*a + 2*b) / b**2\n",
    "print(y)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " - Of what type is the result?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<class 'float'>\n"
     ]
    }
   ],
   "source": [
    "print(type(y))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " - How can you force the result to be a whole number?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1 <class 'int'>\n"
     ]
    }
   ],
   "source": [
    "# Use integer conversion (flooring)\n",
    "y_ = int(y)\n",
    "print(y_, type(y_))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2 <class 'int'>\n"
     ]
    }
   ],
   "source": [
    "# Use round() (closest integer)\n",
    "y_ = round(y)\n",
    "print(y_, type(y_))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " - Of what type is the result if $b = 1$?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "5.0 <class 'float'>\n"
     ]
    }
   ],
   "source": [
    "a = 1\n",
    "b = 1\n",
    "y = (3*a + 2*b) / b**2\n",
    "print(y, type(y))  # Still a float because of division"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " - What happens if $b = 0$?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "ename": "ZeroDivisionError",
     "evalue": "division by zero",
     "output_type": "error",
     "traceback": [
      "\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
      "\u001b[0;31mZeroDivisionError\u001b[0m                         Traceback (most recent call last)",
      "\u001b[0;32m<ipython-input-6-79b38353c79e>\u001b[0m in \u001b[0;36m<module>\u001b[0;34m\u001b[0m\n\u001b[1;32m      1\u001b[0m \u001b[0ma\u001b[0m \u001b[0;34m=\u001b[0m \u001b[0;36m1\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m      2\u001b[0m \u001b[0mb\u001b[0m \u001b[0;34m=\u001b[0m \u001b[0;36m0\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0;32m----> 3\u001b[0;31m \u001b[0my\u001b[0m \u001b[0;34m=\u001b[0m \u001b[0;34m(\u001b[0m\u001b[0;36m3\u001b[0m\u001b[0;34m*\u001b[0m\u001b[0ma\u001b[0m \u001b[0;34m+\u001b[0m \u001b[0;36m2\u001b[0m\u001b[0;34m*\u001b[0m\u001b[0mb\u001b[0m\u001b[0;34m)\u001b[0m \u001b[0;34m/\u001b[0m \u001b[0mb\u001b[0m\u001b[0;34m**\u001b[0m\u001b[0;36m2\u001b[0m  \u001b[0;31m# Raises an error!\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0m\u001b[1;32m      4\u001b[0m \u001b[0mprint\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0my\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mtype\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0my\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n",
      "\u001b[0;31mZeroDivisionError\u001b[0m: division by zero"
     ]
    }
   ],
   "source": [
    "a = 1\n",
    "b = 0\n",
    "y = (3*a + 2*b) / b**2  # Raises an error!\n",
    "print(y, type(y))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "__2__ Calculate the gravitational force between the moon and the earth at a distance of $r = 3.84\\times 10^8$ m, with masses $m_\\mathrm{earth} = 5.97\\times 10^{24}$ kg, $m_\\mathrm{moon} = 7.35\\times 10^{22}$ kg. Use $G = 6.67 \\times 10^{-11}$ m$^3$ kg$^{-1}$ s$^{-2}$ for the gravitational constant."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Gravitational force:  1.9848379516601562e+20 N\n"
     ]
    }
   ],
   "source": [
    "r = 3.84e8\n",
    "m_earth = 5.97e24\n",
    "m_moon = 7.35e22\n",
    "G = 6.67e-11\n",
    "\n",
    "F = (G * m_earth * m_moon) / r**2\n",
    "print(\"Gravitational force: \", F, \"N\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "__Advanced 1__ How does Python handle infinity and absent numeric values? What is\n",
    "the result of a multiplication of these values with 0?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<div class=\"alert alert-info\">\n",
    "\n",
    "**Note:** Advanced exercises are not mandatory to get the points.\n",
    "\n",
    "</div>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "nan\n"
     ]
    }
   ],
   "source": [
    "print(float('nan') * 0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "nan\n"
     ]
    }
   ],
   "source": [
    "print(float('inf') * 0)"
   ]
  }
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