title: If Y is reducible to X in polynomial time, then how is it true that X is at least as hard as Y?
body: <p>I am having difficulty understanding the relationship between the complexity of two classes of problems, say NP-hard and NP-complete problems.</p>
  <p>The answer at <a href="https://stackoverflow.com/a/1857342/">https://stackoverflow.com/a/1857342/</a> states:</p>
  <blockquote>
  <h3>NP Hard</h3>
  <p>Intuitively, these are the problems that are <em>at least as hard as the NP-complete problems</em>. Note that NP-hard problems <em>do not have to be in NP</em>, and <em>they do not have to be decision problems</em>.</p>
  <p>The precise definition here is that <em>a problem <code>X</code> is NP-hard, if there is an NP-complete problem <code>Y</code>, such that <code>Y</code> is reducible to <code>X</code> in polynomial time</em>.</p>
  </blockquote>
  <p>If a problem <code>Y</code> can be reduced to <code>X</code> in polynomial time, should we not say that <code>Y</code> is at least as hard as <code>X</code>? If a problem <code>Y</code> is reducible to <code>X</code> in polynomial time, then the time required to solve <code>Y</code> is polynomial time + the time required to solve <code>X</code>. So it appears to me that problem <code>Y</code> is at least as hard as <code>X</code>.</p>
  <p>But the quoted text above says just the opposite. It says, if an NP-complete problem <code>Y</code> is reducible to an NP-hard problem <code>X</code>, then the NP-hard problem is at least as hard as the NP-complete problem.</p>
  <p>How does this make sense? Where am I making an error in thinking?</p>