relyt
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Find a counterexample to the statement "For all real numbers u and v, (u + v)^2 is not equal to u2 + v2."
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Counterexample: (u + v)^2 ≠ u2 + v2
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What work have you done on the problem? Have you tried anything at all? It's actually harder to find an example than it is a counterexample...
(I assume you meant to write something like [itex](u + v)^2[/itex] -- the proper way to write that when you can't use a superscript is as (u+v)^2, because what you actually wrote is to multiply by 2, and that's a very different problem.
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(I assume you meant to write something like [itex](u + v)^2[/itex] -- the proper way to write that when you can't use a superscript is as (u+v)^2, because what you actually wrote is to multiply by 2, and that's a very different problem.
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relyt
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Hey Hurkyl,
I've tried a couple of things, but I know they are not right. Should I post them here anyway :(
I've tried a couple of things, but I know they are not right. Should I post them here anyway :(
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Definitely. It is forum policy that we won't offer much help until you've shown that you've worked on a problem... (p.s. wee the edits in my previous post)relyt said:
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You want a counterexample for a basic property of Real Numbers? Yes, I know what set of numbers would give a counterexample. It is the set {}.relyt said:
relyt
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symbolipoint said:
Now I'm really confused
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Lol he's just being slight cruel.
Try substitute some small numbers for u and v, see if your Left hand side is equal to your right hand side.
Try substitute some small numbers for u and v, see if your Left hand side is equal to your right hand side.
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No, he's just being completely wrong.
To find a counter example to "(x+ y)2 is NOT x2+ y2", you must use x and y so that (x+y)2= x2+ y2.
Presumably you know that (x+ y)2= x2+ 2xy+ y2. That means you must find x and y so that x2+ 2xy+ y2= x2+ y2.
Notice that the squares on both sides cancel. What does that leave you with?
To find a counter example to "(x+ y)2 is NOT x2+ y2", you must use x and y so that (x+y)2= x2+ y2.
Presumably you know that (x+ y)2= x2+ 2xy+ y2. That means you must find x and y so that x2+ 2xy+ y2= x2+ y2.
Notice that the squares on both sides cancel. What does that leave you with?
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From original: "(u + v)^2 is not equal to u2 + v2."
Actually, I misread the original relation. Either that or the "2" was not shown as exponentiation; but as shown on the right-hand side, the "2" are not shown as exponents.
Actually, I misread the original relation. Either that or the "2" was not shown as exponentiation; but as shown on the right-hand side, the "2" are not shown as exponents.
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No, I read it as "Find a counterexample to 'for all real numbers x,y it is NOT true that [itex](u+ v)^2= u^2+ v^2[/itex]'" and there is an easy counterexample as I pointed out.
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HallsofIvy said:
That is not what was written originally. The 2's on the righthand side were not exponentiated.
Tobias Funke
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The notation isn't the problem. The statement "For all real numbers x,y, (x+y)^2!=2x+2y" is still false. I read it the same as Hurkyl.
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Tobias Funke said:
Why so? try [itex]y=-x[/itex] and [itex]y=2-x[/itex]
edit: I misread "for all real numbers" as "for what real numbers"