What's correct?

Subhotosh Khan said:
Both are correct - test with a = 4 and a=9
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No no! That test is not valid as it doesn't confirm the results for all real numbers. You know this. You are extremely good in math (much better than me!) but you are still a trained engineer and make silly mathematical comments like you just did!
 
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Steven G said:
No no! That test is not valid as it doesn't confirm the results for all real numbers. You know this. You are extremely good in math (much better than me!) but you are still a trained engineer and make silly comments like you just did!
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Testing is not the same as proving.
 
You are correct to be concerned with fractional exponents.
For example in x^(2/4) it allows x to be negative. The reason is if x<0, then x^2>0 and you can now compute (x^2)^(1/4).
Now if you reduce 2/4 to 1/2 you have trouble as x^(1/2) does not allow x<0. Be careful!
 
Steven G said:
You are correct to be concerned with fractional exponents.
For example in x^(2/4) it allows x to be negative. The reason is if x<0, then x^2>0 and you can now compute (x^2)^(1/4).
Now if you reduce 2/4 to 1/2 you have trouble as x^(1/2) does not allow x<0. Be careful!
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Oh dear me. I'm not liking this!
So you are saying that x^(2/4) is not equal to x^(1/2) ??
 
I agree that in \(\displaystyle \sqrt[4]{a^2}\), \(\displaystyle a\) can be negative, but not in \(\displaystyle \sqrt{a}\).
 
Harry_the_cat said:
Yes I agree -when using radical notation. Not sure when using exponent notation though as in post #12.
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I am sure that a^(1/2) does not equal a^(2/4).
Now what do we do with a^.5?? Do we replace .5 = 1/2, 2/4 or 3/6.
Hopefully Dr Peterson can come to the rescue.
 
Steven G said:
I am sure that a^(1/2) does not equal a^(2/4).
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Have you had your coffee this morning?
[imath]\dfrac{1}{2}=\dfrac{2}{4}[/imath] or [imath]a\ge 0\text{ then }\sqrt{a}=\sqrt[4]{a^2}[/imath]
 
Steven G said:
Yes, SK did say to test with a=4 and a=9 but he clearly meant that will serve as a proof for all numbers.
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Steven,

I did not mean and I did not write "proof". My contention is that if an equality holds good for 2 different numbers (other than 0 and 1 - sometimes those behave deceptively) - then the work is "most probably" correct. We must not loose site of FIND. It was not required in the OP to prove the answer "Found" is correct.
 
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