the fraction of a floating point is always &quot;starts with on invisible 1.xxxx&quot; so that forces unique representations for your exponent for any given number.  Is it not clear why this is the case?<p>let&#x27;s take a simple case where you have one bit of fraction.<p><pre><code>   exponent 0, 2^0 = 1, fraction (1)0 = 1.0b = 1

   exponent 0, 2^0 = 1, fraction (1)1 = 1.1b = 1.5

   exponent 1, 2^1 = 2, fraction (1)0 = 10b = 2

   exponent 1, 2^1 = 2, fraction (1)1 = 11b = 3

</code></pre>
if we have more fraction bits it looks like this:<p><pre><code>   exponent 0, 2^0 = 1, fraction (1)00 = 1.00b = 1

   exponent 0, 2^0 = 1, fraction (1)01 = 1.01b = 1.25

   exponent 0, 2^0 = 1, fraction (1)10 = 1.10b = 1.5

   exponent 0, 2^0 = 1, fraction (1)11 = 1.11b = 1.75

   exponent 1, 2^1 = 2, fraction (1)00 = 10.0b = 2

   exponent 1, 2^1 = 2, fraction (1)01 = 10.1b = 2.5

   exponent 1, 2^1 = 2, fraction (1)10 = 11.0b = 3

   exponent 1, 2^1 = 2, fraction (1)11 = 11.1b = 3.5
</code></pre>
So for any given exponent multiplication with a fraction with an invisible bit creates products that are hemmed between 1.0x2^n and 2.0*2^n so there&#x27;s no collisions between any pair (e,f) in your representation.