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Asked by Jayshree Last Modified
Answer To show that 2+22+2
is not a rational number, we'll use a proof by contradiction.
Assume that 2+22+2
is rational. That means it can be expressed as the ratio of two integers aa and bb where b≠0b=0 and aa and bb have no common factors other than 1:
2+2=ab2+2
=ba
Now, let's rearrange this equation to isolate 22
:
2=ab−22
=ba−2
2=a−2bb2
=ba−2b
Now, square both sides:
2=(a−2bb)22=(ba−2b)2
2=(a−2b)2b22=b2(a−2b)2
2b2=(a−2b)22b2=(a−2b)2
2b2=a2−4ab+4b22b2=a2−4ab+4b2
0=a2−4ab+2b20=a2−4ab+2b2
This equation represents a quadratic equation in terms of aa. Now, let's consider this equation modulo 2. This means we'll look at the remainders when dividing each term by 2.
0≡a2−4ab+2b2(mod2)0≡a2−4ab+2b2(mod2)
0≡a2(mod2)0≡a2(mod2)
Since the square of any integer is congruent to either 0 or 1 modulo 2, a2≡0(mod2)a2≡0(mod2) implies that aa itself must be even.
Let a=2ka=2k, where kk is an integer.
Now, substitute a=2ka=2k into the equation:
0=(2k)2−4(2k)b+2b20=(2k)2−4(2k)b+2b2
0=4k2−8kb+2b20=4k2−8kb+2b2
0=2(2k2−4kb+b2)0=2(2k2−4kb+b2)
Since 22 is a prime number, for 2(2k2−4kb+b2)2(2k2−4kb+b2) to be 00, the term inside the parentheses must be divisible by 22. But if 22 divides 2k2−4kb+b22k2−4kb+b2, then 22 divides each of its terms, including b2b2. This implies that bb is also even.
Now, if both aa and bb are even, then they have a common factor of 22, contradicting our initial assumption that aa and bb have no common factors other than 1.
Thus, our initial assumption that 2+22+2
is rational must be false. Therefore, 2+22+2
is irrational.
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