Obvious Theorems (Which Are False)

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[jiaweiliu]

Proof:
0.142857… x 7 =
1/7 x 7 = 1
= 1

[HaldaneSmith]

x = 0.999*
10x = 9.999*

Subract first equation from the second:

9x = 9
x = 1

[b_z]

Of course they’re equal if they weren’t then by the rules of real numbers you would be able to find a number in between them but you can’t so therefore they must be the same.

[janus]

It's quite easy to show that the argument
0.9<1, 0.99<1 and so on therefore 0.99…<1 is wrong, because it would imply:
3<π, 3.1<π, 3.14<π and so on, therefore π is greater than its decimal expansion, which is obviously wrong.

[justyourfriendlyneighborho]

Why is there even a war over this? There are so many simple proofs to this and people still don't seem to understand how infinity works and say it's wrong.

[axelinedgelord]

those are terminating decimals, there’s an obvious difference.

[wscamel]

It makes sense and also it doesn't makes sense.

[TheEternalVortex]

One of the first theorems you prove about limits is: a_n < b_n ==> lim_{n->∞} a_n <= b_n
Note the strict inequality becomes non-strict.

It's pretty intuitive why this happens with many examples, e.g. a_n = 0, b_n = 1/n.

[TopRob]

Anyways
i^i is a pretty interesting problem.
But e^(-pi/2) is not the only solution. It has infinite. I won't explain now but e^(-5pi/2) and e^(3pi/2) also equals i^i

[deadplex]

The way I thought of this is if the difference of and 1 is infinitely small then we can think of their difference as 0 (just the way I thought of it)

[huhr-uzqz]

if x is less than 1, that means you can make it 1 by adding some value.(e.g. 0.9+0.1=1) but when x=0.999…, no matter how small the value is, it will always be bigger than 1. this means there is no gap between 1 and 0.999…, therefore, 1=0.999…

[talkingmurga]

Here is the proof that 0.9... is 1 (I used ... as infinitely repitition after decimal operator) :-

0.9... × 1 = 0.9...

0.9... × 10 = 9.9...

0.9... × 10 - 0.9 × 1 = 9.9... - 0.9...

Or, 0.9...(10 - 1) = 9

Or, 0.9... = 9 / 9 = 1

[HolyCow]

Well infinity is an interesting concept
U cannot intuitively predict it

[N-methylphenylpropan--amine]

New students doing calc 1 try to drag the concept of limits here and convince themselves with statements like "ohh it infinitely gets close to 1 but never is 1"

There is no concept of limits here, it literally doesn't exist in this case. (not with finite number of 9s, rather an infinite 9s) is 1 because the gap between this and 1 is 0. Like you literally won't be able to imagine a distinct 0.9... without falling back to 1.

What IS called "approaching 1" is when your numbers go 0.9865, 0.9986, For limits there isn't an infinite number of 9s, you'll always get a different number after a certain number of 9s. Try coding a program that takes different values of x and prints out 1 - 1/x for them with good precision. Notice how the numbers go. Make the loop as big as possible (maybe 10000)

[vortanoise.]

So:
2a-a=a
And let's say that a=0.9999...
And 2a-1=a because 2*0.9999...=1.999...
Also 2a²-a=a² and this implying that a²=a and in real numbers the only number that squared is equal to itsel is 1.

[prithvisinghpanwar]

0.9999....=x 1
0.9999...×10=10x
2
subtracting equation 1 from 2

9=9x
9/9=x
x=1=0.999....





For some people who might think that in
9 is written n times after the point
so
0.999....×10 - 0.999....
not equal to 9 but

that is not correct because 9 is not written n times
it is written ♾️ times and everyone knows
♾️-1= ♾️
is 9

[itsphoenixingtime]

Something from 3B1B about what being true of the sequence might not be true for the limit. In this case though the sequence implies 0.999(for any natural number digits of 9) < 1, in the limit to infinity they are equal***

***Subject to tons of debate.

[mzhoda]

If you try converting into fraction you will get 1/1 which is just 1.

[josephmax]

Please check your converting method to fraction, it may be wrong

[advaykumar]

What is the greatest integer function of 0.9 repeating?