SOLUTION: True or false (a) If p is rational, then p^2 is rational. (b) If p is irrational, then p^2 is irrational. (c) If p^2 is rational, then p is rational

Algebra ->  Real-numbers -> SOLUTION: True or false (a) If p is rational, then p^2 is rational. (b) If p is irrational, then p^2 is irrational. (c) If p^2 is rational, then p is rational      Log On


   

Question 1209262: True or false
(a) If p is rational, then p^2 is rational.
(b) If p is irrational, then p^2 is irrational.
(c) If p^2 is rational, then p is rational
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Part (a)

Let m and n represent integers where n is nonzero.
p+=+m%2Fn is rational since it's a ratio or fraction of two integers.

p%5E2+=+%28m%2Fn%29%5E2+=+%28m%2Fn%29%2A%28m%2Fn%29+=+%28m%5E2%29%2F%28n%5E2%29 is also rational.

m%5E2 and n%5E2 are integers.
Squaring an integer leads to an integer result.
We conclude statement (a) is always true

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Part (b)

This claim is not true for every possible irrational value p. So we must conclude it is false
The claim is true when p+=+root%283%2C5%29 or p+=+pi for example.

But it's false when p+=+sqrt%287%29. Note that after squaring both sides we get p%5E2+=+7 which is rational.

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Part (c)

This statement appears to be true since it's just the reverse of statement (a)

But we can construct this counter-example
p%5E2+=+3 ---> matrix%281%2C3%2Cp+=+sqrt%283%29%2C%22or%22%2Cp=-sqrt%283%29%29
which are irrational. We cannot write sqrt%283%29 as a ratio of two integers.

So we conclude the statement is false
Sometimes statement (c) would be true (for instance when p%5E2+=+16), but as shown above, it's also sometimes false.

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Answers:
(a) true
(b) false
(c) false