Question

How To Prove Infinity - Infinity = 1

(Suggestions are welcome).

Answer 1

sec^2(theta) -tan^2(theta) = 1

Let theta = 90

sec^2(90) = 1/cos^90 = 1/0 = infinity
tan^2(90) = sin^2(90)/cos^2(90) = 1/0 = infinity

Therefore,
sec^2(90) - tan^2(90) = 1
LHS
sec^2(90) - tan^2(90)
Using the values we found above:
sec^2(90) - tan^2(90) = infinity - infinity = 1

Answer 2

I always thought infinity is not a fixed number (if so, what is it??)
if it's not fixed, you can't find the differences between it and anything else.

Answer 3

I do know infinity is not a fixed number.

I happen to come up with this proof will playing around with some BASIC Trigonometric equations.

Answer 4

Infinity - Infinity = 1

(divide both sides of the
equation by infinity)

1 - 1 = 1 / Infinity
0 = 0

QED

Answer 5

Infinity DOES NOT EQUAL Infinity.

Answer 6

contradiction

Answer 7

How exactly did I contradict myself?

Answer 8

You've shown that
infinity - infinity = 1
aka
infinity = infinity +1

but
substituting this back into the original equation
Infinity - Infinity = 1
(infinity +1) - Infinity = 1
infinity - Infinity +1 = 1
infinity - Infinity = 0
infinity = Infinity

Answer 9

I made a little error in the above post,let me correct it:

sec^2(theta) -tan^2(theta) = 1

Let theta = 90

sec^2(90) = 1/cos^2(90) = 1/0 = infinity
tan^2(90) = sin^2(90)/cos^2(90) = 1/0 = infinity

Therefore,
sec^2(90) - tan^2(90) = 1
LHS
sec^2(90) - tan^2(90)
Using the values we found above:
sec^2(90) - tan^2(90) = infinity - infinity = 1

Answer 10

You must define what you mean by infinity. There are more than one type of infinity, in fact there are infinite.

Consider the real numbers, you can't put these in an indexed list so it is called uncountable infinite where the integers can be listed ...,-3,-2,-1,0,1,2,3,.... and this is called countable infinite.

So to talk about subtraction of infinities we MUST use the definition that it is the cardinality of the two infinite sets.

Subtraction is not defined for infinity in the normal since.

\(|\mathbb{R}|-|\mathbb{Z}|=|\mathbb{R}-\mathbb{Z}|=|\mathbb{R}|\)

but \(|\mathbb{Z}|-|\mathbb{Z}|=|\{\}|=0\)

But you will never find two infinite sets whos "set subtraction" is equal to \(1\).

Answer 11

AMEN

Answer 12

upvotes zzzzzzzzrock

Answer 13

@OP:\[\sin^290^\circ+\cos^290^\circ=1\implies \frac{\sin^290^\circ}{\cos^290^\circ}+\frac{\cos^290^\circ}{\cos^290^\circ}=\frac{1}{\cos^290^\circ}\]...but \(\cos^290^\circ=0\). Your proof is wrong because it relies on the ridiculous notion of dividing by zero, which would also allow you to prove other things that don't make sense like \(1=0\).

For a concrete example, consider the function \(\ln x-\ln(x+1)\). I don't know if you've taken a course in calculus yet, but if you have any working familiarity with the logarithmic function, you should still be able to grasp that as \(x\) gets really big, the value of \(\ln x\) also gets really big (try it on a calculator: what's \(\ln10000\)? \(\ln999999\)? etc). The same goes for \(\ln(x+1)\).

Forgiving the abuse of notation, for very larges values of \(x\), we essentially have \(\ln x=\infty\) and \(\ln(x+1)=\infty\). This suggests \(\ln x-\ln(x+1)=\infty-\infty\), which you claim to be equal to \(1\) by your "proof". Yet the properties of logarithms and polynomial division tell us that \(\ln x-\ln(x+1)=\ln\dfrac{x}{x+1}=\ln\left(1-\dfrac{1}{x+1}\right)\).

You probably realize \(\dfrac{1}{\infty}=0\), so for large \(x\) we get \(\ln\left(1-\dfrac{1}{x+1}\right)=\ln1=0\). So \(\infty-\infty=0\). Both conclusions can't be correct, unless you're fine with working in some bizarro mathematical hellscape where \(1=0\).

Answer 14

\(\dfrac{1}{\infty}\notin \mathbb{R}\text{ and } 0\in \mathbb{R}\)

Answer 15

I see where i went wrong.
Thank you all for the corrections.