u/Archeus__

Let S be an infinite series where S = 1-1+1-1+1-1...

We can group these terms so S = 1+(-1+1)+(-1+1)...

Thus, S = 1+0+0...

We can group S in another way: S = (1-1)+(1-1)+(1-1)...

This way, S = 0+0+0...

In RDM specifically, S = 1 in case 1 and S = 0 in case 2. So, 0=1.

Since RDM treats series as live, you can't argue this series is not divergent.

I don't think this sub should be about arguing 0.(9)=1 because in RDM this might not be true. Instead, it should be about finding inconsistencies like this one.

9.(9) is simply 0.(9)*10 also as said by SPP. Thus, 9.(9)=0.(9)*9. Additionally, 9.(9)-0.(9)=10*0.(9)-0.(9)=9*0.(9) by distributive property. 9.(9)-0.(9) is also equal to 9+0.(9)-0.(9)which is equal to 9. So, 9=9*0.(9). By the fundamentals of algebra, 1=0.(9).

0.(9)/0.(3)=3. Thus, 0.(9)/3=0.(3). We can represent 0.(9) as 1-1/10\^k where k is pushed to the limitless. So, (1-1/10\^k)/3=0.(3). Therefore, 1/3-1/(3\*10\^k)=0.(3.). Since 1/3 = 0.(3), 0.(3)-1/(3\*10\^k)=0.(3). Thus, -1/(3\*10\^k)=0, or 1/10\^k=0. Because 0.(9) is represented by 1-1/10\^k or 1-0, 0.(9)=1.

QED

P.S. SPP, you just deflected my proof last time. Actually disprove it this time.

P.P.S I know my last proof wasn't that good, so hopefully this time it will be better

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0.(9)/0.(3)=3. Thus, 0.(9)/3=0.(3). We can represent 0.(9) as 1-1/10^k where k is pushed to the limitless. So, (1-1/10^k)/3=0.(3). Therefore, 1/3-1/(3*10^k)=0.(3.). Since 1/3 = 0.(3), 0.(3)-1/(3*10^k)=0.(3). Thus, -1/(3*10^k)=0, or 1/10^k=0. Because 0.(9) is represented by 1-1/10^k or 1-0, 0.(9)=1.

QED

P.S. SPP, you just deflected my proof last time. Actually disprove it this time.

P.P.S I know my last proof wasn't that good, so hopefully this time it will be better

SPP, has said he agrees with these three statements: 1-10^n=0.(9) Where n is pushed to the limitless, 1/3=0.(3), and (1-10^n)/(0.(3))=3.

We can substitute the first equation into the third to get (1-10^n)/(1/3)=3. Multiplying both sides by 1/3, we have (1-10^n)/(1/3)*1/3=3*1/3. By "divide negation", we can simplify this to 1-10^n=1. In other words, 0.(9)=1

Qed

Ps actually disprove this instead of referring to your other "proofs".