Disprove- by finding a suitable counter example- the statement x-2-y-2 all x-y-

Question

Toppr Admin · Accepted Answer

For negative number The statement-xA0-x2-gt-y2 For all x -gt- y is disproveIf x-x2212-3 and y-x2212-4 Then -x2212-3-gt-x2212-4But in case -xA0-x2-x2212-3-2-9 and y2-x2212-4-2-16So x2-lt-y2-xA0-Then-xA0-xA0-the statement-xA0-x2-gt-y2 For all x -gt- y is disprove

Disprove, by finding a suitable counter example, the statement $x^{2} > y^{2}$ for all $x > y$ .

For negative number The statement $x^{2} > y^{2}$ For all x > y is disprove
If $x = - 3$ and $y = - 4$ Then $- 3 > - 4$
But in case $x^{2} = (- 3)^{2} = 9$ and $y^{2} = (- 4)^{2} == 16$
So $x^{2} < y^{2}$
Then the statement $x^{2} > y^{2}$ For all x > y is disprove

Disprove, by finding a suitable counter example, the statement x2>y2 for all x>y.

For negative number The statement x2>y2 For all x > y is disproveIf x=−3 and y=−4 Then −3>−4But in case x2=(−3)2=9 and y2=(−4)2==16So x2<y2 Then the statement x2>y2 For all x > y is disprove

Disprove, by finding a suitable counter example, the statement $x^{2} > y^{2}$ for all $x > y$ .

For negative number The statement $x^{2} > y^{2}$ For all x > y is disprove
If $x = - 3$ and $y = - 4$ Then $- 3 > - 4$
But in case $x^{2} = (- 3)^{2} = 9$ and $y^{2} = (- 4)^{2} == 16$
So $x^{2} < y^{2}$
Then the statement $x^{2} > y^{2}$ For all x > y is disprove