Is this Proof correct?

Hi Brilliant Community ,

Recently I was Solving Some Problems and i was perplexed by a expression

and I want to know if this expression is right .

\[\large \dots\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}\] \[\large {= 2^{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \dots}}\] \[\large{= 2 ^{\frac{1}{∞}}}\] \[\large{= 2^{0}}\]
=1\large{= 1}

This means that 2=1\dots\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}=1
21=1\large{ 2 ^{\frac{1}{∞}}=1} 2=1\large{2=1^{∞}}
Is this proof wrong ? and if I am wrong , tell me where it is

(all help appreciated)

Please reshare this note so that we can get better answers .

Note by A Former Brilliant Member
6 years, 6 months ago

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1 vote

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The simple answer is yes, the proof is wrong. In fact, it is correct up until the very end. (or at least the underlying logic is)

The error is between the final two lines. 21=1(21)=12^{\frac{1}{\infty}} = 1 \\ (2^{\frac{1}{\infty}})^{\infty} = 1^{\infty} In this second line, both sides of the equation are indeterminate forms. That is 11.1^{\infty} \neq 1. In fact it is mathematically incorrect really to algebraically put any number to the power of infinity; more properly, you would have to raise it to the power of a limit that approaches infinity. For this reason, to maintain accuracy, you must evaluate the inside exponent first; that is, (21)=1(2^{\frac{1}{\infty}})^{\infty} = 1^{\infty} if a determinate form exists. (edit: to clarify, (21)2(2^{\frac{1}{\infty}})^{\infty} \neq 2)

Caleb Townsend - 6 years, 6 months ago

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I want to correct you a bit

1=11^{\infty } = 1 when 1 on LHS is exact 1

Otherwise 111^{\infty } \neq 1 when 1 is tending 1.

Krishna Sharma - 6 years, 6 months ago

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You are thinking of the limit of 1n1^n as nn approaches infinity. In fact, the number 11^{\infty} itself is indeterminate, therefore not 11 even if the base is exactly 1.1. You could try to define it as the limit I described above, but you could define it as infinitely many other limits as well to get different values. If you don't believe me then check these:

Link 1 Link 2 Link 3

As you can see, wolfram alpha correctly describes it as indeterminate, and the other two links found different values for 1.1^{\infty}. Remember, infinity is not a number; you can't just expect real numbers reasoning to hold at infinity.

Caleb Townsend - 6 years, 6 months ago

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@Caleb Townsend For the first link you can't trust wolfy always, 2nd and 3rd discusses about the the tending 1.

limx0(1+x)1x\displaystyle \lim_{x \to 0} (1+x)^{\frac{1}{x}}

The base in the above example is tending 1.

I'll soon make a note about indeterminant form :)

Krishna Sharma - 6 years, 6 months ago

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@Krishna Sharma Perhaps I can't trust Wolfram Alpha, but I can verify its accuracy in this case. I know for a fact that 11^{\infty} is indeterminate. Infinity is not a number, therefore there is no such thing as 1,1^{\infty}, so the only way to define it is with a limit. If we set the base equal to 11 everywhere and the exponent approaching to infinity, this is still a limit that we are defining and no naturally better than a limit with a base not equal to 1.1.

I guarantee you will not change my mind, but I do recommend that you research this as well; any reliable source will say that 11^{\infty} is indeterminate.

Caleb Townsend - 6 years, 6 months ago

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A very common mistake made is not understanding limits of sequences properly, and thinking that induction applies to "infinity".

As a starting example,
1. Prove that for all nN n \in \mathbb{N} , i=1n(0,1n). \bigcap_{i=1}^n ( 0 , \frac{1}{n} ) \neq \emptyset . 2. Prove that i=1(0,1n)=. \bigcap_{i=1}^ \infty ( 0 , \frac{1}{n} ) = \emptyset .

Calvin Lin Staff - 6 years, 6 months ago

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The problem with your proof is specially in the last two steps. Infinity is not a number and it can not be treated as a number.

Tagesse Lonsamo - 6 years, 4 months ago

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