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What is wrong with this proof?

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i didn't get the mistake in this proof: 2+2=5.Not my proof. Found it on Facebook.

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Note by Priyankar Kumar
3 years, 10 months ago

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There's a subtle mistake in there.

Here's a cryptic hint that might help:

Is \(\sqrt{x^2}\) always equal to \(x\)? Mursalin Habib · 3 years, 10 months ago

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@Mursalin Habib exactly..but the usage of (a-b)^2 somehow disguises this. Priyankar Kumar · 3 years, 10 months ago

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You can't write (4-9/2) as square root of (4-9/2)^2. That is the mistake. Ram Prakash Patel Patel · 3 years, 10 months ago

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@Ram Prakash Patel Patel Why not? Priyankar Kumar · 3 years, 10 months ago

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@Priyankar Kumar Ram Prakash is right. \(4-\frac{9}{2}\) is not equal to \(\sqrt{(4-\frac{9}{2})^2}\). This is what I hinted at in my initial comment. Mursalin Habib · 3 years, 10 months ago

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@Priyankar Kumar Since \(4-9/2\) is negative, while \(\sqrt{(4-9/2)^2}\) is positive. Michael Tang · 3 years, 10 months ago

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@Michael Tang ok..thanks..i get the point. Thanks to Mursalin too. Priyankar Kumar · 3 years, 10 months ago

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@Priyankar Kumar You're welcome! Mursalin Habib · 3 years, 10 months ago

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@Mursalin Habib Look at the last. Assuming the mistake that it is \(-\sqrt{(5-\frac{9}{2})^{2}}\) we will get \(-5+\frac{9}{2}+\frac{9}{2}=4\)So, it is proved that \(2+2=4\) and not \(5\) Fahad Shihab · 3 years, 10 months ago

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\( \sqrt{(4 - \frac{9}{2}) ^{2}} \) is not equal to \( 4 - \frac{9}{2}\) Tan Li Xuan · 3 years, 10 months ago

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@Tan Li Xuan thanks Priyankar Kumar · 3 years, 10 months ago

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\( \sqrt x^2 = |x| \) not \( x\). Lokesh Sharma · 3 years, 10 months ago

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You cant make a negative number positive by just squaring it and taking root. I'm talking about 2nd step, (4-(9/2)) Chirag Pachori · 3 years, 10 months ago

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@Chirag Pachori thanks Priyankar Kumar · 3 years, 10 months ago

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\((4-\frac{9}{2})=-\frac{1}{2}\)

but,

\((5-\frac{9}{2})=\frac{1}{2}\)

in this proof we wrote--

\(2+2=4-\frac{9}{2}+\frac{9}{2}...........step (i)\)

\(=\sqrt{(5-\frac{9}{2})^2}+\frac{9}{2}..........step(viii)\)

\(=5-\frac{9}{2}+\frac{9}{2}..........step(ix)\)

so,what we actually doing is

\(-\frac{1}{2}+\frac{9}{2}=\frac{1}{2}+\frac{9}{2}\)..............consider step (i) and (ix)

this is where we made mistake.

So in this case at the time of removing square root of \( \sqrt{(5-\frac{9}{2})^2}\)

we need to consider \(\sqrt{(5-\frac{9}{2})^2}=-(5-\frac{9}{2})\) Gypsy Singer · 3 years, 10 months ago

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@Gypsy Singer \( \sqrt {(5-\frac{9}{2})^{2}} \) is not equal to \( -(5-\frac{9}{2})\) Tan Li Xuan · 3 years, 10 months ago

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@Gypsy Singer ok..i understood thanks. Priyankar Kumar · 3 years, 10 months ago

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@Priyankar Kumar Careful! Gypsy's last statement is incorrect.

\(\sqrt{(5-\frac{9}{2})^2}\) is not equal to \(-(5-\frac{9}{2})\).

However, \(\sqrt{(4-\frac{9}{2})^2}=-(4-\frac{9}{2})\). Mursalin Habib · 3 years, 10 months ago

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I think so that the square roots can't be split... Fahad Shihab · 3 years, 10 months ago

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It is wrong because \(\sqrt{x^2}\) is not always equal to \(x\). It has two values, that are \(x\) and \(-x\) Akshat Jain · 3 years, 10 months ago

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@Akshat Jain You should be careful about what you write. \(\sqrt{x^2}\) [the principal square-root of \(x^2\)] does not have two values.

Take this for example: what is \(\sqrt{(1-\sqrt{3})^2}\). Is it \((1-\sqrt{3})\)? Or is it \(-(1-\sqrt{3})\)? Or is it both?

According to you, both of these should be correct. Are they?

Also see the solutions for this problem where almost all of them make this same mistake. Mursalin Habib · 3 years, 10 months ago

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@Mursalin Habib Mmm, that got me. Akshat Jain · 3 years, 10 months ago

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@Akshat Jain sqrt( x^2 ) is always | x | Vinayak K · 3 years, 10 months ago

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@Akshat Jain exactly..but in the subsequent steps, (a-b)^2 disguises this. Priyankar Kumar · 3 years, 10 months ago

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Where is your proof ?? Toan Pham Quang · 3 years, 10 months ago

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@Toan Pham Quang it's an external link. Click on the heading of this discussion. Priyankar Kumar · 3 years, 10 months ago

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Did you see the term: 2 x 4 x 9/2? if we cancel 2 then the answer will be 39 but when we multiply the numerators, it is equal to 29. So maybe it is the mistake James Vincent Llandelar · 3 years, 10 months ago

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@James Vincent Llandelar No,the calculation is\( 2 \times 4 \times \frac{9}{2}\).If we cancel the 2's,then it becomes \( 4 \times 9 =36\) and if we multiply the numerators it is \( \frac{72}{2} = 36\). So this is not the mistake. Tan Li Xuan · 3 years, 10 months ago

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@James Vincent Llandelar No sir, i don't think so. Priyankar Kumar · 3 years, 10 months ago

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