Does 0.999 Equal 1? Maths Project

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The discussion confirms that 0.99999... is mathematically equal to 1, emphasizing this as a mathematical identity rather than a matter of opinion. Participants mention the importance of significant figures in understanding this concept. The equivalence is likened to the limit of 1/x approaching 0 as x approaches infinity. The thread also notes that similar questions have been previously posted by the user. The conversation concludes with a reminder that this is a well-established mathematical fact.
chickenguy
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HI epople, i am doing a project on maths and i would like to know this first of all. i am telling you that 0.99999...=1, do you think this is correct or incorrect and do you have any special opinions?
 
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Ask your teacher about significant figures.
It's correct to the extent that you reduce the number of sigfigs
 
Yes they are equal, just like the limit for x \rightarrow \infty of 1/x is 0.
Just remember that it is not a matter of opinion or polling, but a mathematical identity!
 
You've already posted several threads that are virtually identical to this one.

Thread closed.

- Warren
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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