Question 1209262
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Part (a)


Let m and n represent integers where n is nonzero.
{{{p = m/n}}} is <font color=red>ratio</font>nal since it's a <font color=red>ratio</font> or fraction of two integers.


{{{p^2 = (m/n)^2 = (m/n)*(m/n) = (m^2)/(n^2)}}} is also rational.


{{{m^2}}} and {{{n^2}}} are integers. 
Squaring an integer leads to an integer result.
We conclude statement (a) is <font color=red>always true</font>


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Part (b)


This claim is not true for every possible irrational value p. So we must conclude it is <font color=red>false</font>
The claim is true when {{{p = root(3,5)}}} or {{{p = pi}}} for example.


But it's false when {{{p = sqrt(7)}}}. Note that after squaring both sides we get {{{p^2 = 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^2 = 3}}} ---> {{{matrix(1,3,p = sqrt(3),"or",p=-sqrt(3))}}}
which are irrational. We cannot write {{{sqrt(3)}}} as a ratio of two integers.


So we conclude the statement is <font color=red>false</font>
Sometimes statement (c) would be true (for instance when {{{p^2 = 16}}}), but as shown above, it's also sometimes false.


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Answers:
(a) <font color=red>true</font>
(b) <font color=red>false</font>
(c) <font color=red>false</font>
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