-0.000 000 000 742 147 676 541 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 541(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
-0.000 000 000 742 147 676 541(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:

|-0.000 000 000 742 147 676 541| = 0.000 000 000 742 147 676 541


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: 0.000 000 000 742 147 676 541.

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) 0.000 000 000 742 147 676 541 × 2 = 0 + 0.000 000 001 484 295 353 082;
  • 2) 0.000 000 001 484 295 353 082 × 2 = 0 + 0.000 000 002 968 590 706 164;
  • 3) 0.000 000 002 968 590 706 164 × 2 = 0 + 0.000 000 005 937 181 412 328;
  • 4) 0.000 000 005 937 181 412 328 × 2 = 0 + 0.000 000 011 874 362 824 656;
  • 5) 0.000 000 011 874 362 824 656 × 2 = 0 + 0.000 000 023 748 725 649 312;
  • 6) 0.000 000 023 748 725 649 312 × 2 = 0 + 0.000 000 047 497 451 298 624;
  • 7) 0.000 000 047 497 451 298 624 × 2 = 0 + 0.000 000 094 994 902 597 248;
  • 8) 0.000 000 094 994 902 597 248 × 2 = 0 + 0.000 000 189 989 805 194 496;
  • 9) 0.000 000 189 989 805 194 496 × 2 = 0 + 0.000 000 379 979 610 388 992;
  • 10) 0.000 000 379 979 610 388 992 × 2 = 0 + 0.000 000 759 959 220 777 984;
  • 11) 0.000 000 759 959 220 777 984 × 2 = 0 + 0.000 001 519 918 441 555 968;
  • 12) 0.000 001 519 918 441 555 968 × 2 = 0 + 0.000 003 039 836 883 111 936;
  • 13) 0.000 003 039 836 883 111 936 × 2 = 0 + 0.000 006 079 673 766 223 872;
  • 14) 0.000 006 079 673 766 223 872 × 2 = 0 + 0.000 012 159 347 532 447 744;
  • 15) 0.000 012 159 347 532 447 744 × 2 = 0 + 0.000 024 318 695 064 895 488;
  • 16) 0.000 024 318 695 064 895 488 × 2 = 0 + 0.000 048 637 390 129 790 976;
  • 17) 0.000 048 637 390 129 790 976 × 2 = 0 + 0.000 097 274 780 259 581 952;
  • 18) 0.000 097 274 780 259 581 952 × 2 = 0 + 0.000 194 549 560 519 163 904;
  • 19) 0.000 194 549 560 519 163 904 × 2 = 0 + 0.000 389 099 121 038 327 808;
  • 20) 0.000 389 099 121 038 327 808 × 2 = 0 + 0.000 778 198 242 076 655 616;
  • 21) 0.000 778 198 242 076 655 616 × 2 = 0 + 0.001 556 396 484 153 311 232;
  • 22) 0.001 556 396 484 153 311 232 × 2 = 0 + 0.003 112 792 968 306 622 464;
  • 23) 0.003 112 792 968 306 622 464 × 2 = 0 + 0.006 225 585 936 613 244 928;
  • 24) 0.006 225 585 936 613 244 928 × 2 = 0 + 0.012 451 171 873 226 489 856;
  • 25) 0.012 451 171 873 226 489 856 × 2 = 0 + 0.024 902 343 746 452 979 712;
  • 26) 0.024 902 343 746 452 979 712 × 2 = 0 + 0.049 804 687 492 905 959 424;
  • 27) 0.049 804 687 492 905 959 424 × 2 = 0 + 0.099 609 374 985 811 918 848;
  • 28) 0.099 609 374 985 811 918 848 × 2 = 0 + 0.199 218 749 971 623 837 696;
  • 29) 0.199 218 749 971 623 837 696 × 2 = 0 + 0.398 437 499 943 247 675 392;
  • 30) 0.398 437 499 943 247 675 392 × 2 = 0 + 0.796 874 999 886 495 350 784;
  • 31) 0.796 874 999 886 495 350 784 × 2 = 1 + 0.593 749 999 772 990 701 568;
  • 32) 0.593 749 999 772 990 701 568 × 2 = 1 + 0.187 499 999 545 981 403 136;
  • 33) 0.187 499 999 545 981 403 136 × 2 = 0 + 0.374 999 999 091 962 806 272;
  • 34) 0.374 999 999 091 962 806 272 × 2 = 0 + 0.749 999 998 183 925 612 544;
  • 35) 0.749 999 998 183 925 612 544 × 2 = 1 + 0.499 999 996 367 851 225 088;
  • 36) 0.499 999 996 367 851 225 088 × 2 = 0 + 0.999 999 992 735 702 450 176;
  • 37) 0.999 999 992 735 702 450 176 × 2 = 1 + 0.999 999 985 471 404 900 352;
  • 38) 0.999 999 985 471 404 900 352 × 2 = 1 + 0.999 999 970 942 809 800 704;
  • 39) 0.999 999 970 942 809 800 704 × 2 = 1 + 0.999 999 941 885 619 601 408;
  • 40) 0.999 999 941 885 619 601 408 × 2 = 1 + 0.999 999 883 771 239 202 816;
  • 41) 0.999 999 883 771 239 202 816 × 2 = 1 + 0.999 999 767 542 478 405 632;
  • 42) 0.999 999 767 542 478 405 632 × 2 = 1 + 0.999 999 535 084 956 811 264;
  • 43) 0.999 999 535 084 956 811 264 × 2 = 1 + 0.999 999 070 169 913 622 528;
  • 44) 0.999 999 070 169 913 622 528 × 2 = 1 + 0.999 998 140 339 827 245 056;
  • 45) 0.999 998 140 339 827 245 056 × 2 = 1 + 0.999 996 280 679 654 490 112;
  • 46) 0.999 996 280 679 654 490 112 × 2 = 1 + 0.999 992 561 359 308 980 224;
  • 47) 0.999 992 561 359 308 980 224 × 2 = 1 + 0.999 985 122 718 617 960 448;
  • 48) 0.999 985 122 718 617 960 448 × 2 = 1 + 0.999 970 245 437 235 920 896;
  • 49) 0.999 970 245 437 235 920 896 × 2 = 1 + 0.999 940 490 874 471 841 792;
  • 50) 0.999 940 490 874 471 841 792 × 2 = 1 + 0.999 880 981 748 943 683 584;
  • 51) 0.999 880 981 748 943 683 584 × 2 = 1 + 0.999 761 963 497 887 367 168;
  • 52) 0.999 761 963 497 887 367 168 × 2 = 1 + 0.999 523 926 995 774 734 336;
  • 53) 0.999 523 926 995 774 734 336 × 2 = 1 + 0.999 047 853 991 549 468 672;
  • 54) 0.999 047 853 991 549 468 672 × 2 = 1 + 0.998 095 707 983 098 937 344;

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:


0.000 000 000 742 147 676 541(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 541(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 541(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 676 541 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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