Prove or disprove that if a and b are rational numbers, then ab is also rational.

Question

aprovard · Accepted Answer

Given statement: If a and b are rational numbers, then $a^{b}$ is also rational.
The given statement is false, which will be disproven using a counterexample.
Counterexample
If $a = - 1$ and $b = \frac{1}{2}$ , we then obtain ${(- 1)}^{\frac{1}{2}} = \sqrt{- 1}$ which is not rational number (it is called an imaginary number) and thus $a^{b}$ is not necessarily rational.
Result:
Statement is false, one counterexample is for a=-1 and $b = \frac{1}{2}$ .

fudzisako · Accepted Answer

To prove or disprove that if a and b are rational numbers, then ab is also rational, let&#039;s consider a counterexample.Counterexample:Let a=2 and b=12. Both a and b are rational numbers. Now, let&#039;s calculate ab.ab=(2)12=2Since 2 is an irrational number, 2 is also irrational. Therefore, we have found a counterexample where a and b are rational numbers, but ab is irrational.Hence, we have disproven the statement that if a and b are rational numbers, then ab is also rational.

Jazz Frenia · Accepted Answer

Let a=2 and b=12, both of which are rational numbers.Now, let&#039;s evaluate ab:ab=212We know that 2 is an irrational number. If 212 were rational, then it would contradict the irrationality of 2. Therefore, we can conclude that if a and b are rational numbers, ab is not necessarily a rational number.Hence, we have disproven the statement.

Andre BalkonE · Accepted Answer

Step 1:Proof:Let a and b be rational numbers, such that a=pq and b=mn, where p, q, m, and n are integers with q≠0 and n≠0. We want to show that ab is also rational.Consider ab=(pq)mn. Using the property of exponentiation, we can rewrite this expression as ((pq)1n)m.Since 1n is also a rational number, let 1n=rs, where r and s are integers with s≠0.Substituting this into the expression above, we have ((pq)rs)m.By the definition of rational exponentiation, (pq)rs is also a rational number. Let (pq)rs=xy, where x and y are integers with y≠0.Substituting this back into the expression, we have (xy)m. Again, using the property of exponentiation, we have (xmym).Since xm and ym are both integers, ab=xmym is a rational number.Therefore, we have proven that if a and b are rational numbers, then ab is also rational.Step 2:Counterexample:Let a=23 and b=2. Here, a is a rational number, but b is an irrational number.If we calculate ab, we get (23)2. It is known that (23)2 is an irrational number.Hence, we have found a counterexample where a is a rational number, b is an irrational number, but ab is an irrational number. This disproves the statement.Therefore, we can conclude that the statement &#039;&#039;if a and b are rational numbers, then ab is also rational&#039;&#039; is false.

Prove or disprove that if a and b are rational numbers, then a^{b} is also

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