# ## MIT License

# Permission is hereby granted, free of charge, to any person obtaining a copy
# of this software and associated documentation files (the "Software"), to deal
# in the Software without restriction, including without limitation the rights
# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
# copies of the Software, and to permit persons to whom the Software is
# furnished to do so, subject to the following conditions:
#
# The above copyright notice and this permission notice shall be included in
# all copies or substantial portions of the Software.
#
# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
# AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
# THE SOFTWARE.
#

# In[5]:
import numpy as np
# Let's make two numbers with very similar magnitude:
# In[6]:
x = 1.48234
y = 1.48235
# Now let's compute their difference in double precision:
# In[7]:
x_dbl = np.float64(x)
y_dbl = np.float64(y)
diff_dbl = x_dbl-y_dbl
print(repr(diff_dbl))
# * What would the correct result be?
# * What has happened here?
# -------------
# Can you predict what will happen in single precision?
# In[8]:
x_sng = np.float32(x)
y_sng = np.float32(y)
diff_sng = x_sng-y_sng
print(diff_sng)
# In[ ]: