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Can we represent $\sqrt{2}$ exactly even with infinite bits in mantissa in floating point notation or otherwise. We actually have to prove this is not possible. But why can't we if we have infinite bits? Infinite bits means the binary representation of the mantissa can be indefinitely large. The mantissa is in IEEE 754 32 bit format and in a hypothetical world where we don't have a bound on the number of bits which can be used to represent it.

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    $\begingroup$ It is unclear what you mean by "infinite bits". What does that mean? The mantissa is a whole, integer number. This necessarily means it must have a finite number of bits. $\endgroup$ – Tom van der Zanden Aug 27 at 14:39
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    $\begingroup$ The universe is finite and incapable of storing anything infinite. $\endgroup$ – gnasher729 Aug 27 at 14:52
  • $\begingroup$ It's not clear what you mean by either "represent exactly" or "represent with infinite[ly many] bits". For any real number $r\in[0,1)$, let $a_1=0$ if $r<1/2$ and $a_1=1$, otherwise. Having defined $a_1, \dots, a_n$, let $a_{n+1}=0$ if $r-\sum_{i=1}^n a_i2^{-i}<2^{-(n+1)}$ and $a_{n+1}=1$, otherwise. Then $\lim_{n\to\infty}\sum_{i=1}^n a_i2^{-i}=r$. $\endgroup$ – David Richerby Aug 27 at 15:51
  • $\begingroup$ Note that real numbers are usually defined as a limit of finite sums. If you allow for sums with infinite number of terms directly (as your infinite representation would require), then arithmetics become inconsistent. $\endgroup$ – Dmitri Urbanowicz Aug 28 at 12:38