if 1+1 =2 then 2+2=5

 

its true, but how?

it isn`t true, so there is no "How"

if its true is a null hypothesis, so feel free to make up your own answer

Having said that of course, I don`t suppose your "Friend" could try and convince my Bank Manager of this "Fact" by any chance????

mine too. that would be awesome. just keep applying the logic and your balance will increase geometrically.

You can probably "show" it using what is known as a fallacy. That is some apparently correct derivation which upon closer inspection uses some dubious logic.

 

Most common examples involve division by zero. Indeed using division by zero you can show that any number is equal to any other!

 

But as you know division by zero is not allowed!

if 1+1 =2 then 2+2=5

 

its true, but how?

 

Some values of 2 are bigger than others.

2+2!=5 either.

if 1+1 =2 then 2+2=5

 

its true, but how?

how would 2+2=5. i learnt those facts in kindergarten (before primary) (elementary for the states)

 

how is it true?

 

i honestly dont get it

 

 

[edit] I am sick of merging your multiple consecutive posts. Either think before you hit the submit button, or learn how to edit. Sayo.

It isn't true, don't worry about it.

I guess it can be "shown" to be true, by using some logical mistakes along the way, as has already been shown. I wish they taught me about this stuff at school, so I would know about how basic truths are derived from axioms and be able to apply the logical process involved. It is one of the many neglected areas of mathematical education IMO.

I guess it can be "shown" to be true, by using some logical mistakes along the way, as has already been shown.

 

Try this one.

 

I can "prove" 1 = 2

 

[math]a=b[/math] by assumption

multipliy by b

[math]a^{2} = ab[/math]

minus [math]b^{2}[/math] form both sides.

[math]a^{2}-b^{2} = b(a-b)[/math]

using the difference of two squares identity.

[math](a+b)(a-b) = b(a-b)[/math]

divide both sides by [math](a-b)[/math]

[math]a+b = b[/math]

 

Now use our initial assumption and we get 1= 2.

 

Can you spot my mistake? If you know this trick already don't give it away, let the others have a think about it.

 

I am sure there are many many other similar fallacies out there. Maybe we should start a new thread where people can post them?

ahh that one is a classic. i looked a bit of an idiot for believing that one when i was in 2nd year of secondary school.

Here's a pretty standard one:

[math]1^1=1^0[/math]

Take the log of the equation:

[math]ln1^1=ln1^0[/math]

Equivilantly,

[math]1ln1=0ln1[/math]

[math]1=0[/math]

 

This one is a little tougher.

The imaginary number, [math]i=\sqrt{-1}[/math]

Now, let [math]-1=-1[/math]

We can also write this as

[math]\frac{-1}{1}=\frac{1}{-1}[/math]

Take the square root,

[math]\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}[/math]

[math]\frac{i}{1}=\frac{1}{i}[/math]

Since by definiton, [math]i^2=-1[/math] then

[math]1=-1[/math]

[math]1ln1=0ln1[/math]

[math]1=0[/math]

 

ln1 = 0. Can't divide by 0.

I suspect he knew that, he wasn't suggesting that those tricks had any truth behind them.

I suspect he knew that, he wasn't suggesting that those tricks had any truth behind them.

 

I know, I'm just solving his challenge.

editted

Try this one.

 

I can "prove" 1 = 2

 

[math]a=b[/math] by assumption

multipliy by b

[math]a^{2} = ab[/math]

minus [math]b^{2}[/math] form both sides.

[math]a^{2}-b^{2} = b(a-b)[/math]

using the difference of two squares identity.

[math](a+b)(a-b) = b(a-b)[/math]

divide both sides by [math](a-b)[/math]

[math]a+b = b[/math]

 

Now use our initial assumption and we get 1= 2.

 

Can you spot my mistake? If you know this trick already don't give it away, let the others have a think about it.

 

I am sure there are many many other similar fallacies out there. Maybe we should start a new thread where people can post them?

 

[hide]Pretty sure the mistake (or at least a mistake) here is dividing by (a-b) which we all know, since a=b is the same as dividing by (a-a) or (b-b) which is dividing by 0.[/hide]

 

Also, one possible rationalization for the OP: [math]1+1=2[/math] is equal to 1 more than [math]1^2=1[/math], therefore [math]a+a=a^2+1[/math] therefore [math]2+2=2^2+1=4+1=5[/math].

 

We all know, of course, that that isn't true, but I felt like pointing it out. I'm tired and bored, so it comes to that.

how would 2+2=5. i learnt those facts in kindergarten (before primary) (elementary for the states)

 

how is it true?

 

i honestly dont get it

 

 

[edit] I am sick of merging your multiple consecutive posts. Either think before you hit the submit button, or learn how to edit. Sayo.

 

"how in the hell do you plan to fail"

what do u mean by plan to fail?

This one is a little tougher.

The imaginary number, [math]i=\sqrt{-1}[/math]

Now, let [math]-1=-1[/math]

We can also write this as

[math]\frac{-1}{1}=\frac{1}{-1}[/math]

Take the square root,

[math]\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}[/math]

[math]\frac{i}{1}=\frac{1}{i}[/math]

Since by definiton, [math]i^2=-1[/math] then

[math]1=-1[/math]

 

I'm stumped, what is the trick behind this one?

 

Edit: Is it order of operations? PEMDAS here in America, BEDMANS in Canada, I don't know other countries' convention names. But anyway before you do the exponent of 1/2 on both sides (aka take the square root) you have to complete the division first! Because you're taking the squareroot of each side in its entirety, you have to perform the operation of each side first.

The square root law

[math]

\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}

[/math]

only applies for real positive values of x and y.

 

It should actually be

[math]

\sqrt{\frac{x}{y}}=\pm\frac{\sqrt{x}}{\sqrt{y}}

[/math]

if u have a software that was instructed to round up figures,,, than 2+2 could be equal to 5.

 

2.4+2.4 = 4.8

the output would be displayed as 2+2 = 5 and it would even be considered as ture.

apart from that i cant think of a way by which it can be logically stated that its ture at the moment.