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# What is wrong with this proof?

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 $$x$$? · 3 years, 10 months ago

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

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

Why not? · 3 years, 10 months ago

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. · 3 years, 10 months ago

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

ok..thanks..i get the point. Thanks to Mursalin too. · 3 years, 10 months ago

You're welcome! · 3 years, 10 months ago

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 $$5$$ · 3 years, 10 months ago

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

thanks · 3 years, 10 months ago

$$\sqrt x^2 = |x|$$ not $$x$$. · 3 years, 10 months ago

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

thanks · 3 years, 10 months ago

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

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 $$\sqrt{(5-\frac{9}{2})^2}=-(5-\frac{9}{2})$$ · 3 years, 10 months ago

$$\sqrt {(5-\frac{9}{2})^{2}}$$ is not equal to $$-(5-\frac{9}{2})$$ · 3 years, 10 months ago

ok..i understood thanks. · 3 years, 10 months ago

Careful! Gypsy's last statement is incorrect.

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

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

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

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

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. · 3 years, 10 months ago

Mmm, that got me. · 3 years, 10 months ago

sqrt( x^2 ) is always | x | · 3 years, 10 months ago

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

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

it's an external link. Click on the heading of this discussion. · 3 years, 10 months ago

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 · 3 years, 10 months ago

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. · 3 years, 10 months ago