1.111 000 111 010 100 111 100 100 100 111 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.111 000 111 010 100 111 100 100 100 111 63(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
1.111 000 111 010 100 111 100 100 100 111 63(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 1.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.111 000 111 010 100 111 100 100 100 111 63.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.111 000 111 010 100 111 100 100 100 111 63 × 2 = 0 + 0.222 000 222 020 200 222 200 200 200 223 26;
  • 2) 0.222 000 222 020 200 222 200 200 200 223 26 × 2 = 0 + 0.444 000 444 040 400 444 400 400 400 446 52;
  • 3) 0.444 000 444 040 400 444 400 400 400 446 52 × 2 = 0 + 0.888 000 888 080 800 888 800 800 800 893 04;
  • 4) 0.888 000 888 080 800 888 800 800 800 893 04 × 2 = 1 + 0.776 001 776 161 601 777 601 601 601 786 08;
  • 5) 0.776 001 776 161 601 777 601 601 601 786 08 × 2 = 1 + 0.552 003 552 323 203 555 203 203 203 572 16;
  • 6) 0.552 003 552 323 203 555 203 203 203 572 16 × 2 = 1 + 0.104 007 104 646 407 110 406 406 407 144 32;
  • 7) 0.104 007 104 646 407 110 406 406 407 144 32 × 2 = 0 + 0.208 014 209 292 814 220 812 812 814 288 64;
  • 8) 0.208 014 209 292 814 220 812 812 814 288 64 × 2 = 0 + 0.416 028 418 585 628 441 625 625 628 577 28;
  • 9) 0.416 028 418 585 628 441 625 625 628 577 28 × 2 = 0 + 0.832 056 837 171 256 883 251 251 257 154 56;
  • 10) 0.832 056 837 171 256 883 251 251 257 154 56 × 2 = 1 + 0.664 113 674 342 513 766 502 502 514 309 12;
  • 11) 0.664 113 674 342 513 766 502 502 514 309 12 × 2 = 1 + 0.328 227 348 685 027 533 005 005 028 618 24;
  • 12) 0.328 227 348 685 027 533 005 005 028 618 24 × 2 = 0 + 0.656 454 697 370 055 066 010 010 057 236 48;
  • 13) 0.656 454 697 370 055 066 010 010 057 236 48 × 2 = 1 + 0.312 909 394 740 110 132 020 020 114 472 96;
  • 14) 0.312 909 394 740 110 132 020 020 114 472 96 × 2 = 0 + 0.625 818 789 480 220 264 040 040 228 945 92;
  • 15) 0.625 818 789 480 220 264 040 040 228 945 92 × 2 = 1 + 0.251 637 578 960 440 528 080 080 457 891 84;
  • 16) 0.251 637 578 960 440 528 080 080 457 891 84 × 2 = 0 + 0.503 275 157 920 881 056 160 160 915 783 68;
  • 17) 0.503 275 157 920 881 056 160 160 915 783 68 × 2 = 1 + 0.006 550 315 841 762 112 320 321 831 567 36;
  • 18) 0.006 550 315 841 762 112 320 321 831 567 36 × 2 = 0 + 0.013 100 631 683 524 224 640 643 663 134 72;
  • 19) 0.013 100 631 683 524 224 640 643 663 134 72 × 2 = 0 + 0.026 201 263 367 048 449 281 287 326 269 44;
  • 20) 0.026 201 263 367 048 449 281 287 326 269 44 × 2 = 0 + 0.052 402 526 734 096 898 562 574 652 538 88;
  • 21) 0.052 402 526 734 096 898 562 574 652 538 88 × 2 = 0 + 0.104 805 053 468 193 797 125 149 305 077 76;
  • 22) 0.104 805 053 468 193 797 125 149 305 077 76 × 2 = 0 + 0.209 610 106 936 387 594 250 298 610 155 52;
  • 23) 0.209 610 106 936 387 594 250 298 610 155 52 × 2 = 0 + 0.419 220 213 872 775 188 500 597 220 311 04;
  • 24) 0.419 220 213 872 775 188 500 597 220 311 04 × 2 = 0 + 0.838 440 427 745 550 377 001 194 440 622 08;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.111 000 111 010 100 111 100 100 100 111 63(10) =

0.0001 1100 0110 1010 1000 0000(2)

5. Positive number before normalization:

1.111 000 111 010 100 111 100 100 100 111 63(10) =

1.0001 1100 0110 1010 1000 0000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.111 000 111 010 100 111 100 100 100 111 63(10) =

1.0001 1100 0110 1010 1000 0000(2) =

1.0001 1100 0110 1010 1000 0000(2) × 20


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.0001 1100 0110 1010 1000 0000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

127(10) =

0111 1111(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 000 1110 0011 0101 0100 0000 0 =

000 1110 0011 0101 0100 0000


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0111 1111

Mantissa (23 bits) =
000 1110 0011 0101 0100 0000


Decimal number 1.111 000 111 010 100 111 100 100 100 111 63 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 000 1110 0011 0101 0100 0000

Share

» Convert decimal 1.111 000 111 010 100 111 100 100 100 111 27 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. Gemini Notebook YouTube Video of 8.15 minutes

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111