-0.000 000 000 742 147 508 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 508(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
-0.000 000 000 742 147 508(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:

|-0.000 000 000 742 147 508| = 0.000 000 000 742 147 508


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: 0.000 000 000 742 147 508.

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) 0.000 000 000 742 147 508 × 2 = 0 + 0.000 000 001 484 295 016;
  • 2) 0.000 000 001 484 295 016 × 2 = 0 + 0.000 000 002 968 590 032;
  • 3) 0.000 000 002 968 590 032 × 2 = 0 + 0.000 000 005 937 180 064;
  • 4) 0.000 000 005 937 180 064 × 2 = 0 + 0.000 000 011 874 360 128;
  • 5) 0.000 000 011 874 360 128 × 2 = 0 + 0.000 000 023 748 720 256;
  • 6) 0.000 000 023 748 720 256 × 2 = 0 + 0.000 000 047 497 440 512;
  • 7) 0.000 000 047 497 440 512 × 2 = 0 + 0.000 000 094 994 881 024;
  • 8) 0.000 000 094 994 881 024 × 2 = 0 + 0.000 000 189 989 762 048;
  • 9) 0.000 000 189 989 762 048 × 2 = 0 + 0.000 000 379 979 524 096;
  • 10) 0.000 000 379 979 524 096 × 2 = 0 + 0.000 000 759 959 048 192;
  • 11) 0.000 000 759 959 048 192 × 2 = 0 + 0.000 001 519 918 096 384;
  • 12) 0.000 001 519 918 096 384 × 2 = 0 + 0.000 003 039 836 192 768;
  • 13) 0.000 003 039 836 192 768 × 2 = 0 + 0.000 006 079 672 385 536;
  • 14) 0.000 006 079 672 385 536 × 2 = 0 + 0.000 012 159 344 771 072;
  • 15) 0.000 012 159 344 771 072 × 2 = 0 + 0.000 024 318 689 542 144;
  • 16) 0.000 024 318 689 542 144 × 2 = 0 + 0.000 048 637 379 084 288;
  • 17) 0.000 048 637 379 084 288 × 2 = 0 + 0.000 097 274 758 168 576;
  • 18) 0.000 097 274 758 168 576 × 2 = 0 + 0.000 194 549 516 337 152;
  • 19) 0.000 194 549 516 337 152 × 2 = 0 + 0.000 389 099 032 674 304;
  • 20) 0.000 389 099 032 674 304 × 2 = 0 + 0.000 778 198 065 348 608;
  • 21) 0.000 778 198 065 348 608 × 2 = 0 + 0.001 556 396 130 697 216;
  • 22) 0.001 556 396 130 697 216 × 2 = 0 + 0.003 112 792 261 394 432;
  • 23) 0.003 112 792 261 394 432 × 2 = 0 + 0.006 225 584 522 788 864;
  • 24) 0.006 225 584 522 788 864 × 2 = 0 + 0.012 451 169 045 577 728;
  • 25) 0.012 451 169 045 577 728 × 2 = 0 + 0.024 902 338 091 155 456;
  • 26) 0.024 902 338 091 155 456 × 2 = 0 + 0.049 804 676 182 310 912;
  • 27) 0.049 804 676 182 310 912 × 2 = 0 + 0.099 609 352 364 621 824;
  • 28) 0.099 609 352 364 621 824 × 2 = 0 + 0.199 218 704 729 243 648;
  • 29) 0.199 218 704 729 243 648 × 2 = 0 + 0.398 437 409 458 487 296;
  • 30) 0.398 437 409 458 487 296 × 2 = 0 + 0.796 874 818 916 974 592;
  • 31) 0.796 874 818 916 974 592 × 2 = 1 + 0.593 749 637 833 949 184;
  • 32) 0.593 749 637 833 949 184 × 2 = 1 + 0.187 499 275 667 898 368;
  • 33) 0.187 499 275 667 898 368 × 2 = 0 + 0.374 998 551 335 796 736;
  • 34) 0.374 998 551 335 796 736 × 2 = 0 + 0.749 997 102 671 593 472;
  • 35) 0.749 997 102 671 593 472 × 2 = 1 + 0.499 994 205 343 186 944;
  • 36) 0.499 994 205 343 186 944 × 2 = 0 + 0.999 988 410 686 373 888;
  • 37) 0.999 988 410 686 373 888 × 2 = 1 + 0.999 976 821 372 747 776;
  • 38) 0.999 976 821 372 747 776 × 2 = 1 + 0.999 953 642 745 495 552;
  • 39) 0.999 953 642 745 495 552 × 2 = 1 + 0.999 907 285 490 991 104;
  • 40) 0.999 907 285 490 991 104 × 2 = 1 + 0.999 814 570 981 982 208;
  • 41) 0.999 814 570 981 982 208 × 2 = 1 + 0.999 629 141 963 964 416;
  • 42) 0.999 629 141 963 964 416 × 2 = 1 + 0.999 258 283 927 928 832;
  • 43) 0.999 258 283 927 928 832 × 2 = 1 + 0.998 516 567 855 857 664;
  • 44) 0.998 516 567 855 857 664 × 2 = 1 + 0.997 033 135 711 715 328;
  • 45) 0.997 033 135 711 715 328 × 2 = 1 + 0.994 066 271 423 430 656;
  • 46) 0.994 066 271 423 430 656 × 2 = 1 + 0.988 132 542 846 861 312;
  • 47) 0.988 132 542 846 861 312 × 2 = 1 + 0.976 265 085 693 722 624;
  • 48) 0.976 265 085 693 722 624 × 2 = 1 + 0.952 530 171 387 445 248;
  • 49) 0.952 530 171 387 445 248 × 2 = 1 + 0.905 060 342 774 890 496;
  • 50) 0.905 060 342 774 890 496 × 2 = 1 + 0.810 120 685 549 780 992;
  • 51) 0.810 120 685 549 780 992 × 2 = 1 + 0.620 241 371 099 561 984;
  • 52) 0.620 241 371 099 561 984 × 2 = 1 + 0.240 482 742 199 123 968;
  • 53) 0.240 482 742 199 123 968 × 2 = 0 + 0.480 965 484 398 247 936;
  • 54) 0.480 965 484 398 247 936 × 2 = 0 + 0.961 930 968 796 495 872;

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:


0.000 000 000 742 147 508(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 508(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(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:


0.000 000 000 742 147 508(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 100


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 1100 =

100 1011 1111 1111 1111 1100


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 1100


Decimal number -0.000 000 000 742 147 508 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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