Math, asked by Cms, 10 months ago

if a and b are any two integers such that a>b, then a^2>b^2 . True or false? Give reason.

Answers

Answered by mr165154
2

Answer:

Yes Because It Is Given That,

a>b.

Therefore, Now Squaring Both Sides.

That Is, a²>b².

Hope This Will Help u Mate.


Cms: But it's false...
Answered by brunoconti
0

Answer:

Step-by-step explanation:

NOT true if a and b are integers.

Example: - 1 > - 2 , but 1 < 4 , where 1 = (- 1)^2 and 4 = (- 2)^2

However it is true if a and b are natural numbers.

a > b ---> a - b > 0

---> (a - b)(a + b) > 0

---> a^2 - b^2 > 0

---> a^2 > b^2


Cms: I didn't understand your answer from -- however it is true if a and b are natural numbers
Cms: a-b>0 then how (a-b)(a+b)>0 . Please tell me mate...
brunoconti: this is the formal proof for natural numbers because if a and b are natural numbers such that a > b, then a- b > 0 and a + b > 0 and therefore (a - b)(a + b) > 0 and consequently a^2 - b^2 > 0 and finally a^2 > b^2
brunoconti: BRAINLIEST please
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