M missc New member Joined Mar 28, 2011 Messages 5 Mar 28, 2011 #1 If a > b is a² > b² ? If a > b is a² always greater than b² ? Explain why and give examples. I believe it is, but I'm a bit unsure. Help?
If a > b is a² > b² ? If a > b is a² always greater than b² ? Explain why and give examples. I believe it is, but I'm a bit unsure. Help?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Mar 28, 2011 #2 Re: If a > b is a² > b² ? How about if a and b are negative. Say, \(\displaystyle a=-2, \;\ b=-3\) In this case, is \(\displaystyle a^{2}>b^{2}\)?.
Re: If a > b is a² > b² ? How about if a and b are negative. Say, \(\displaystyle a=-2, \;\ b=-3\) In this case, is \(\displaystyle a^{2}>b^{2}\)?.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Mar 28, 2011 #3 Re: If a > b is a² > b² ? Hello, missc! Did you try some specific examples? \(\displaystyle \text{If }a > b\text{, is }a^2\text{ always greater than }b^2\,?\) \(\displaystyle \text{Explain why and give examples.}\) Click to expand... If negative numbers can be used, the answer is no. \(\displaystyle 2 \,>\, \text{-}3,\;\text{ but . . .}\)
Re: If a > b is a² > b² ? Hello, missc! Did you try some specific examples? \(\displaystyle \text{If }a > b\text{, is }a^2\text{ always greater than }b^2\,?\) \(\displaystyle \text{Explain why and give examples.}\) Click to expand... If negative numbers can be used, the answer is no. \(\displaystyle 2 \,>\, \text{-}3,\;\text{ but . . .}\)
