Define hypothesis math term:
The term hypothesis refers the condition statement just after the word if. In the given conditional statement.
For example Suppose the condition is “in the given quadrilateral figure - four sides are equal in measure then the shape is said to be square” the hypothesis of the given statement is “all four sides are equal in measure. In the given example,” all four sides are equal in measure. “Always, never or sometimes true.
Options:
1)Always
2)Never
3)Sometimes.
Answer is (3)
Solution:
Answer 3) sometimes it is correct because, it’s true if only the given shape is square, In case of other shapes like rectangle, rhombus then condition is not satisfied.
Example 1)
Find the math term hypothesis of the given equation.
If x = 1 then x2 +3x +5 = 9
Option:
1)True
2)false
Answer : 1)True.
Solution:
Substitute x value in the given equation,
X+3x+5 =9
1+3(1) +5 = 9
1+3+5 =9
9=9
Therefore the given condition is true.
Example 2:
Find the math term hypothesis in t he given statement is “Every integer greater than 1 is product of prime number “
Solution:
Let us consider the if x is true statement .
Therefore ,
X`=>` y is a true statement and y is a true conditional statement.
Now,we need to find the series of the statement x1,x2 …..etc.,and then verifies the given conditional statement” Every integer greater than 1 is product of prime number”.
X=x1
X1=x2,
X2=x3,
….
….
Xn-1 =xn
Xn =Y are ture.
Because assume that the given statement X is true then all repeated statement of the simple rule Y is true..
Example 3:
The conditional statement is “if n is an integer then n2 is even and then n is odd.”
Solution:
N is odd number .
so n-1 is even number .
Therefore , we have,
n-1 =2k
where ,k is the constant .
n-1 = 2k ,k `in` Z
n=2k+1
Similarly we have by hypothesis,
n2 =x; x `in` Z
And then,
n2 = 4k2 +4k+1
=2(2k2 +2k)+1
In the above equation, 2(2k2+2k) is even number.
So 2(2k2+2k) +1 is the odd number.Therefore our hypothesis is violated.
So the integer number “n “ must be even number.
So,
X`=>` S =” n2 is even ”
Y `=>` ” n is even ”
Not –Y `=>` ” n is odd ”
Practice problems related to the term hypothesis in math :
1) Prove that if m and n are even integers, then n +m is even.
2) Prove that if n is an odd integer, then n 2 is an odd integer.
The term hypothesis refers the condition statement just after the word if. In the given conditional statement.
For example Suppose the condition is “in the given quadrilateral figure - four sides are equal in measure then the shape is said to be square” the hypothesis of the given statement is “all four sides are equal in measure. In the given example,” all four sides are equal in measure. “Always, never or sometimes true.
Options:
1)Always
2)Never
3)Sometimes.
Answer is (3)
Solution:
Answer 3) sometimes it is correct because, it’s true if only the given shape is square, In case of other shapes like rectangle, rhombus then condition is not satisfied.
Example 1)
Find the math term hypothesis of the given equation.
If x = 1 then x2 +3x +5 = 9
Option:
1)True
2)false
Answer : 1)True.
Solution:
Substitute x value in the given equation,
X+3x+5 =9
1+3(1) +5 = 9
1+3+5 =9
9=9
Therefore the given condition is true.
Example 2:
Find the math term hypothesis in t he given statement is “Every integer greater than 1 is product of prime number “
Solution:
Let us consider the if x is true statement .
Therefore ,
X`=>` y is a true statement and y is a true conditional statement.
Now,we need to find the series of the statement x1,x2 …..etc.,and then verifies the given conditional statement” Every integer greater than 1 is product of prime number”.
X=x1
X1=x2,
X2=x3,
….
….
Xn-1 =xn
Xn =Y are ture.
Because assume that the given statement X is true then all repeated statement of the simple rule Y is true..
Example 3:
The conditional statement is “if n is an integer then n2 is even and then n is odd.”
Solution:
N is odd number .
so n-1 is even number .
Therefore , we have,
n-1 =2k
where ,k is the constant .
n-1 = 2k ,k `in` Z
n=2k+1
Similarly we have by hypothesis,
n2 =x; x `in` Z
And then,
n2 = 4k2 +4k+1
=2(2k2 +2k)+1
In the above equation, 2(2k2+2k) is even number.
So 2(2k2+2k) +1 is the odd number.Therefore our hypothesis is violated.
So the integer number “n “ must be even number.
So,
X`=>` S =” n2 is even ”
Y `=>` ” n is even ”
Not –Y `=>` ” n is odd ”
Practice problems related to the term hypothesis in math :
1) Prove that if m and n are even integers, then n +m is even.
2) Prove that if n is an odd integer, then n 2 is an odd integer.
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