since square root of $2$ is a irrational number, which we know or assume its decimal part is not a finite number, or doesn't terminate, how come we say that this infinite number (not in terms of being a large number) times itself is equals $2$. It is like saying $0.928041\dots$ times itself is some finite number (like $1.64$ not going $3$ dots in the end).
how come square root of 2 times itself equals 2
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It's by definition. $\sqrt 2$ is the unique positive number that, when squared, results in $2$. Are you also surprised that $\frac13+\frac13+\frac13=1$ even though you are adding a number repeatedly to itself that has a decimal representation that is non-terminating? (Similar to above, $\frac13$ is defined as the unique number that, when multiplied by $3$, results in $1$)
Do not confuse a number with the representation of a number by a decimal expansion.
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In the real world, you only want to be accurate up to a point. Within an inch, or a millimetre, or much less if you are building a computer chip.
Two people might agree that two numbers are close enough to being equal, but a third might disagree.
If two numbers are so close that everyone agrees they are close enough - so they agree to every decimal point - then we say they are equal. So $1/3=0.333333333...$
There is no such thing as an infinite number. If you mean that the decimal expansion of $\sqrt2$ never ends, then you are right. But that also applies to the decimal expansion of $\frac13$; however, I suspect that you see no problem with the assertion that $3\times\frac13=1$.
The fact that the decimal expansion of $\sqrt2$ never ends is something that concerns the specific way that we use to represent numbers, which was developed centuries after square roots started to be studied. It is not something about the number $\sqrt2$ itself, which is defined as the only real number greater than or equal to $0$ whose square is $2$.
After having defined it this way, we can try to compute its decimal expansion. For instance, $1^2<2<2^2$, and therefore $1<\sqrt2<2$. Furthermore, $\left(\frac75\right)^2<2<\left(\frac85\right)^2$ and therefore $\frac75<\sqrt2<\frac85(1.4<\sqrt2<1.6)$.