-0.000 000 000 742 117 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 117 4(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
-0.000 000 000 742 117 4(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. Start with the positive version of the number:

|-0.000 000 000 742 117 4| = 0.000 000 000 742 117 4


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 117 4.

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) 0.000 000 000 742 117 4 × 2 = 0 + 0.000 000 001 484 234 8;
  • 2) 0.000 000 001 484 234 8 × 2 = 0 + 0.000 000 002 968 469 6;
  • 3) 0.000 000 002 968 469 6 × 2 = 0 + 0.000 000 005 936 939 2;
  • 4) 0.000 000 005 936 939 2 × 2 = 0 + 0.000 000 011 873 878 4;
  • 5) 0.000 000 011 873 878 4 × 2 = 0 + 0.000 000 023 747 756 8;
  • 6) 0.000 000 023 747 756 8 × 2 = 0 + 0.000 000 047 495 513 6;
  • 7) 0.000 000 047 495 513 6 × 2 = 0 + 0.000 000 094 991 027 2;
  • 8) 0.000 000 094 991 027 2 × 2 = 0 + 0.000 000 189 982 054 4;
  • 9) 0.000 000 189 982 054 4 × 2 = 0 + 0.000 000 379 964 108 8;
  • 10) 0.000 000 379 964 108 8 × 2 = 0 + 0.000 000 759 928 217 6;
  • 11) 0.000 000 759 928 217 6 × 2 = 0 + 0.000 001 519 856 435 2;
  • 12) 0.000 001 519 856 435 2 × 2 = 0 + 0.000 003 039 712 870 4;
  • 13) 0.000 003 039 712 870 4 × 2 = 0 + 0.000 006 079 425 740 8;
  • 14) 0.000 006 079 425 740 8 × 2 = 0 + 0.000 012 158 851 481 6;
  • 15) 0.000 012 158 851 481 6 × 2 = 0 + 0.000 024 317 702 963 2;
  • 16) 0.000 024 317 702 963 2 × 2 = 0 + 0.000 048 635 405 926 4;
  • 17) 0.000 048 635 405 926 4 × 2 = 0 + 0.000 097 270 811 852 8;
  • 18) 0.000 097 270 811 852 8 × 2 = 0 + 0.000 194 541 623 705 6;
  • 19) 0.000 194 541 623 705 6 × 2 = 0 + 0.000 389 083 247 411 2;
  • 20) 0.000 389 083 247 411 2 × 2 = 0 + 0.000 778 166 494 822 4;
  • 21) 0.000 778 166 494 822 4 × 2 = 0 + 0.001 556 332 989 644 8;
  • 22) 0.001 556 332 989 644 8 × 2 = 0 + 0.003 112 665 979 289 6;
  • 23) 0.003 112 665 979 289 6 × 2 = 0 + 0.006 225 331 958 579 2;
  • 24) 0.006 225 331 958 579 2 × 2 = 0 + 0.012 450 663 917 158 4;
  • 25) 0.012 450 663 917 158 4 × 2 = 0 + 0.024 901 327 834 316 8;
  • 26) 0.024 901 327 834 316 8 × 2 = 0 + 0.049 802 655 668 633 6;
  • 27) 0.049 802 655 668 633 6 × 2 = 0 + 0.099 605 311 337 267 2;
  • 28) 0.099 605 311 337 267 2 × 2 = 0 + 0.199 210 622 674 534 4;
  • 29) 0.199 210 622 674 534 4 × 2 = 0 + 0.398 421 245 349 068 8;
  • 30) 0.398 421 245 349 068 8 × 2 = 0 + 0.796 842 490 698 137 6;
  • 31) 0.796 842 490 698 137 6 × 2 = 1 + 0.593 684 981 396 275 2;
  • 32) 0.593 684 981 396 275 2 × 2 = 1 + 0.187 369 962 792 550 4;
  • 33) 0.187 369 962 792 550 4 × 2 = 0 + 0.374 739 925 585 100 8;
  • 34) 0.374 739 925 585 100 8 × 2 = 0 + 0.749 479 851 170 201 6;
  • 35) 0.749 479 851 170 201 6 × 2 = 1 + 0.498 959 702 340 403 2;
  • 36) 0.498 959 702 340 403 2 × 2 = 0 + 0.997 919 404 680 806 4;
  • 37) 0.997 919 404 680 806 4 × 2 = 1 + 0.995 838 809 361 612 8;
  • 38) 0.995 838 809 361 612 8 × 2 = 1 + 0.991 677 618 723 225 6;
  • 39) 0.991 677 618 723 225 6 × 2 = 1 + 0.983 355 237 446 451 2;
  • 40) 0.983 355 237 446 451 2 × 2 = 1 + 0.966 710 474 892 902 4;
  • 41) 0.966 710 474 892 902 4 × 2 = 1 + 0.933 420 949 785 804 8;
  • 42) 0.933 420 949 785 804 8 × 2 = 1 + 0.866 841 899 571 609 6;
  • 43) 0.866 841 899 571 609 6 × 2 = 1 + 0.733 683 799 143 219 2;
  • 44) 0.733 683 799 143 219 2 × 2 = 1 + 0.467 367 598 286 438 4;
  • 45) 0.467 367 598 286 438 4 × 2 = 0 + 0.934 735 196 572 876 8;
  • 46) 0.934 735 196 572 876 8 × 2 = 1 + 0.869 470 393 145 753 6;
  • 47) 0.869 470 393 145 753 6 × 2 = 1 + 0.738 940 786 291 507 2;
  • 48) 0.738 940 786 291 507 2 × 2 = 1 + 0.477 881 572 583 014 4;
  • 49) 0.477 881 572 583 014 4 × 2 = 0 + 0.955 763 145 166 028 8;
  • 50) 0.955 763 145 166 028 8 × 2 = 1 + 0.911 526 290 332 057 6;
  • 51) 0.911 526 290 332 057 6 × 2 = 1 + 0.823 052 580 664 115 2;
  • 52) 0.823 052 580 664 115 2 × 2 = 1 + 0.646 105 161 328 230 4;
  • 53) 0.646 105 161 328 230 4 × 2 = 1 + 0.292 210 322 656 460 8;
  • 54) 0.292 210 322 656 460 8 × 2 = 0 + 0.584 420 645 312 921 6;

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).


5. 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:


0.000 000 000 742 117 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0111 10(2)

6. Positive number before normalization:

0.000 000 000 742 117 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 117 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 0111 0111 10(2) × 20 =

1.1001 0111 1111 1011 1011 110(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1011 1011 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1101 1101 1110 =

100 1011 1111 1101 1101 1110


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

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1101 1101 1110


Decimal number -0.000 000 000 742 117 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1101 1101 1110

Share

» Convert decimal -0.000 000 000 742 123 8 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 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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