I'm supposed to evaluate the proof, edit if there are errors or confirm that it is correct (b) Proposition. For all nonzero integers a and b, if a + 2b # 3 and9a + 2b 1, then the equation ax +2bx = 3 does not have asolution that is a natural number.Proof. We will use a proof by contradiction. Let us assume that thereexist nonzero integers a and b such that a + 26 = 3 and 9a + 2b = 1and an + 2bn = 3, where n is a natural number. First, we will solveone equation for 2b; doing this, we obtaina + 26 = 326 = 3– a.(1)We can now substitute for 2b in an + 2bn = 3. This givesan + (3-a)n =3an +3n-an = 3n (an? +3-a)= 3.(2)By the closure properties of the integers. (an+3-4) is an integerand, hence, equation (2) implics that divides B. So n= 1 or TnWhen we substitute n = 1 into the equation an +2bn = 3, we obtaina+2h = 3. This is a contrădiction since we are toldinthe propositionthat a + 2h # 3. This proves that the negation of the proposition isfalse and, hence, the proposition is true.12:49 AM

Name: I'm supposed to evaluate the proof, edit if there are errors or confir
Uploaded: 2024-03-20T06:38:44.000Z
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Description: I'm supposed to evaluate the proof, edit if there are errors or confirm that it is correct (b) Proposition. For all nonzero integers a and b, if a + 2b # 3 and9a + 2b 1, then the equation ax +2bx = 3 does not have asolution that is a natural number.P