user115192: I do a sort of bitcoin 101 presentation for folks where I work. In one of the recent sessions, someone asked why the proof of work difficulty requires a minimum number of leading 0s. Why not 1s or something else?
Not understanding the math behind it in-depth, the best answer I can come up with is that the protocol requires that the proof of work hash (hash of nonce and block data) should be equal to or lower than the difficulty target. Hence, you can interpret it as having to contain a certain number of leading 0s.
Is there a better explanation?
Thanks!

user48884: 
Not understanding the math behind it in-depth, the best answer I can come up with is that the protocol requires that the proof of work hash (hash of nonce and block data) should be equal to or lower than the difficulty target. Hence, you can interpret it as having to contain a certain number of leading 0s.

This is exactly what it is. The actual calculation has nothing to do with leading 0s. It's entirely just a comparison of integers, which has the effect of producing leading 0s.

115192: <p>I do a sort of bitcoin 101 presentation for folks where I work. In one of the recent sessions, someone asked why the proof of work difficulty requires a minimum number of leading 0s. Why not 1s or something else?</p>
<p>Not understanding the math behind it in-depth, the best answer I can come up with is that the protocol requires that the proof of work hash (hash of nonce and block data) should be equal to or lower than the difficulty target. Hence, you can interpret it as having to contain a certain number of leading 0s.</p>
<p>Is there a better explanation?</p>
<p>Thanks!</p>
 48884: <blockquote>
<p>Not understanding the math behind it in-depth, the best answer I can come up with is that the protocol requires that the proof of work hash (hash of nonce and block data) should be equal to or lower than the difficulty target. Hence, you can interpret it as having to contain a certain number of leading 0s.</p>
</blockquote>
<p>This is exactly what it is. The actual calculation has nothing to do with leading 0s. It's entirely just a comparison of integers, which has the effect of producing leading 0s.</p>
