SOLUTION: please show the conjecture is false by coming up with a counterexample. the square root of a number x is always less than x.

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Question 206509: please show the conjecture is false by coming up with a counterexample.

the square root of a number x is always less than x.
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(28476) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%29%3Cx Start with the given inequality.


To prove this false with a counter-example, we need to use a small number (that is less than 1). I'm going to use 1%2F4 (since 1%2F4%3C1)


sqrt%281%2F4%29%3C1%2F4 Plug in x=1%2F4


sqrt%281%29%2Fsqrt%284%29%3C1%2F4 Break up the root.


1%2Fsqrt%284%29%3C1%2F4 Evaluate the square root of 1 to get 1.


1%2F2%3C1%2F4 Evaluate the square root of 4 to get 2.


1%2A4%3C1%2A2 Cross multiply (this will help us determine which side is larger)


4%3C2 Multiply


Since the inequality is FALSE, this means that 1%2F2%3C1%2F4 is FALSE (it turns out that one-half is actually larger than a quarter...draw out a picture to verify yourself). So this means that sqrt%281%2F4%29%3C1%2F4 is also false.

So we've shown that the inequality sqrt%28x%29%3Cx is false for all real numbers.

Note: it turns out that sqrt%28x%29%3Cx is only true if x%3E1

Answer by stanbon(57214) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt(1/4) = 1/2
and 1/2 is greater than 1/4
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Cheers,
Stan H.