Show \(f(x)=x-[x]\) is Neither One-One Nor Onto

📺 Video Explanation

📝 Question

Show that:

\[ f:\mathbb{R}\to\mathbb{R},\quad f(x)=x-[x] \]

is neither one-one nor onto.

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✅ Solution

🔹 Step 1: Understand Function

Here:

\[ [x] \]

means greatest integer less than or equal to \(x\).

So:

\[ f(x)=x-[x] \]

is fractional part of \(x\).

Thus:

\[ 0\leq f(x)<1 \]

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🔹 Step 2: Check One-One

Take:

\[ x=1.5,\quad x=2.5 \]

Then:

\[ f(1.5)=1.5-1=0.5 \]

\[ f(2.5)=2.5-2=0.5 \]

Different inputs give same output.

❌ Not one-one.

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🔹 Step 3: Check Onto

Range:

\[ [0,1) \]

But codomain:

\[ \mathbb{R} \]

Values like:

\[ 2,\ -1,\ 1.5 \]

are not attained.

❌ Not onto.

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🎯 Final Answer

\[ \boxed{\text{f(x)=x-[x] is neither one-one nor onto}} \]

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🚀 Exam Shortcut

• Fractional part repeats after every integer
• Range is always between 0 and 1
• So neither injective nor surjective