-0.000 000 000 742 147 383 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 383(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
-0.000 000 000 742 147 383(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:

|-0.000 000 000 742 147 383| = 0.000 000 000 742 147 383


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: 0.000 000 000 742 147 383.

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) 0.000 000 000 742 147 383 × 2 = 0 + 0.000 000 001 484 294 766;
  • 2) 0.000 000 001 484 294 766 × 2 = 0 + 0.000 000 002 968 589 532;
  • 3) 0.000 000 002 968 589 532 × 2 = 0 + 0.000 000 005 937 179 064;
  • 4) 0.000 000 005 937 179 064 × 2 = 0 + 0.000 000 011 874 358 128;
  • 5) 0.000 000 011 874 358 128 × 2 = 0 + 0.000 000 023 748 716 256;
  • 6) 0.000 000 023 748 716 256 × 2 = 0 + 0.000 000 047 497 432 512;
  • 7) 0.000 000 047 497 432 512 × 2 = 0 + 0.000 000 094 994 865 024;
  • 8) 0.000 000 094 994 865 024 × 2 = 0 + 0.000 000 189 989 730 048;
  • 9) 0.000 000 189 989 730 048 × 2 = 0 + 0.000 000 379 979 460 096;
  • 10) 0.000 000 379 979 460 096 × 2 = 0 + 0.000 000 759 958 920 192;
  • 11) 0.000 000 759 958 920 192 × 2 = 0 + 0.000 001 519 917 840 384;
  • 12) 0.000 001 519 917 840 384 × 2 = 0 + 0.000 003 039 835 680 768;
  • 13) 0.000 003 039 835 680 768 × 2 = 0 + 0.000 006 079 671 361 536;
  • 14) 0.000 006 079 671 361 536 × 2 = 0 + 0.000 012 159 342 723 072;
  • 15) 0.000 012 159 342 723 072 × 2 = 0 + 0.000 024 318 685 446 144;
  • 16) 0.000 024 318 685 446 144 × 2 = 0 + 0.000 048 637 370 892 288;
  • 17) 0.000 048 637 370 892 288 × 2 = 0 + 0.000 097 274 741 784 576;
  • 18) 0.000 097 274 741 784 576 × 2 = 0 + 0.000 194 549 483 569 152;
  • 19) 0.000 194 549 483 569 152 × 2 = 0 + 0.000 389 098 967 138 304;
  • 20) 0.000 389 098 967 138 304 × 2 = 0 + 0.000 778 197 934 276 608;
  • 21) 0.000 778 197 934 276 608 × 2 = 0 + 0.001 556 395 868 553 216;
  • 22) 0.001 556 395 868 553 216 × 2 = 0 + 0.003 112 791 737 106 432;
  • 23) 0.003 112 791 737 106 432 × 2 = 0 + 0.006 225 583 474 212 864;
  • 24) 0.006 225 583 474 212 864 × 2 = 0 + 0.012 451 166 948 425 728;
  • 25) 0.012 451 166 948 425 728 × 2 = 0 + 0.024 902 333 896 851 456;
  • 26) 0.024 902 333 896 851 456 × 2 = 0 + 0.049 804 667 793 702 912;
  • 27) 0.049 804 667 793 702 912 × 2 = 0 + 0.099 609 335 587 405 824;
  • 28) 0.099 609 335 587 405 824 × 2 = 0 + 0.199 218 671 174 811 648;
  • 29) 0.199 218 671 174 811 648 × 2 = 0 + 0.398 437 342 349 623 296;
  • 30) 0.398 437 342 349 623 296 × 2 = 0 + 0.796 874 684 699 246 592;
  • 31) 0.796 874 684 699 246 592 × 2 = 1 + 0.593 749 369 398 493 184;
  • 32) 0.593 749 369 398 493 184 × 2 = 1 + 0.187 498 738 796 986 368;
  • 33) 0.187 498 738 796 986 368 × 2 = 0 + 0.374 997 477 593 972 736;
  • 34) 0.374 997 477 593 972 736 × 2 = 0 + 0.749 994 955 187 945 472;
  • 35) 0.749 994 955 187 945 472 × 2 = 1 + 0.499 989 910 375 890 944;
  • 36) 0.499 989 910 375 890 944 × 2 = 0 + 0.999 979 820 751 781 888;
  • 37) 0.999 979 820 751 781 888 × 2 = 1 + 0.999 959 641 503 563 776;
  • 38) 0.999 959 641 503 563 776 × 2 = 1 + 0.999 919 283 007 127 552;
  • 39) 0.999 919 283 007 127 552 × 2 = 1 + 0.999 838 566 014 255 104;
  • 40) 0.999 838 566 014 255 104 × 2 = 1 + 0.999 677 132 028 510 208;
  • 41) 0.999 677 132 028 510 208 × 2 = 1 + 0.999 354 264 057 020 416;
  • 42) 0.999 354 264 057 020 416 × 2 = 1 + 0.998 708 528 114 040 832;
  • 43) 0.998 708 528 114 040 832 × 2 = 1 + 0.997 417 056 228 081 664;
  • 44) 0.997 417 056 228 081 664 × 2 = 1 + 0.994 834 112 456 163 328;
  • 45) 0.994 834 112 456 163 328 × 2 = 1 + 0.989 668 224 912 326 656;
  • 46) 0.989 668 224 912 326 656 × 2 = 1 + 0.979 336 449 824 653 312;
  • 47) 0.979 336 449 824 653 312 × 2 = 1 + 0.958 672 899 649 306 624;
  • 48) 0.958 672 899 649 306 624 × 2 = 1 + 0.917 345 799 298 613 248;
  • 49) 0.917 345 799 298 613 248 × 2 = 1 + 0.834 691 598 597 226 496;
  • 50) 0.834 691 598 597 226 496 × 2 = 1 + 0.669 383 197 194 452 992;
  • 51) 0.669 383 197 194 452 992 × 2 = 1 + 0.338 766 394 388 905 984;
  • 52) 0.338 766 394 388 905 984 × 2 = 0 + 0.677 532 788 777 811 968;
  • 53) 0.677 532 788 777 811 968 × 2 = 1 + 0.355 065 577 555 623 936;
  • 54) 0.355 065 577 555 623 936 × 2 = 0 + 0.710 131 155 111 247 872;

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:


0.000 000 000 742 147 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) × 20 =

1.1001 0111 1111 1111 1111 010(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 1111 1111 010


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 1111 1111 1010 =

100 1011 1111 1111 1111 1010


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 1111 1111 1010


Decimal number -0.000 000 000 742 147 383 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1010

Share

» Convert decimal -0.000 000 000 742 147 433 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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