-0.000 000 000 742 147 676 392 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 392(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
-0.000 000 000 742 147 676 392(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:

|-0.000 000 000 742 147 676 392| = 0.000 000 000 742 147 676 392


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: 0.000 000 000 742 147 676 392.

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) 0.000 000 000 742 147 676 392 × 2 = 0 + 0.000 000 001 484 295 352 784;
  • 2) 0.000 000 001 484 295 352 784 × 2 = 0 + 0.000 000 002 968 590 705 568;
  • 3) 0.000 000 002 968 590 705 568 × 2 = 0 + 0.000 000 005 937 181 411 136;
  • 4) 0.000 000 005 937 181 411 136 × 2 = 0 + 0.000 000 011 874 362 822 272;
  • 5) 0.000 000 011 874 362 822 272 × 2 = 0 + 0.000 000 023 748 725 644 544;
  • 6) 0.000 000 023 748 725 644 544 × 2 = 0 + 0.000 000 047 497 451 289 088;
  • 7) 0.000 000 047 497 451 289 088 × 2 = 0 + 0.000 000 094 994 902 578 176;
  • 8) 0.000 000 094 994 902 578 176 × 2 = 0 + 0.000 000 189 989 805 156 352;
  • 9) 0.000 000 189 989 805 156 352 × 2 = 0 + 0.000 000 379 979 610 312 704;
  • 10) 0.000 000 379 979 610 312 704 × 2 = 0 + 0.000 000 759 959 220 625 408;
  • 11) 0.000 000 759 959 220 625 408 × 2 = 0 + 0.000 001 519 918 441 250 816;
  • 12) 0.000 001 519 918 441 250 816 × 2 = 0 + 0.000 003 039 836 882 501 632;
  • 13) 0.000 003 039 836 882 501 632 × 2 = 0 + 0.000 006 079 673 765 003 264;
  • 14) 0.000 006 079 673 765 003 264 × 2 = 0 + 0.000 012 159 347 530 006 528;
  • 15) 0.000 012 159 347 530 006 528 × 2 = 0 + 0.000 024 318 695 060 013 056;
  • 16) 0.000 024 318 695 060 013 056 × 2 = 0 + 0.000 048 637 390 120 026 112;
  • 17) 0.000 048 637 390 120 026 112 × 2 = 0 + 0.000 097 274 780 240 052 224;
  • 18) 0.000 097 274 780 240 052 224 × 2 = 0 + 0.000 194 549 560 480 104 448;
  • 19) 0.000 194 549 560 480 104 448 × 2 = 0 + 0.000 389 099 120 960 208 896;
  • 20) 0.000 389 099 120 960 208 896 × 2 = 0 + 0.000 778 198 241 920 417 792;
  • 21) 0.000 778 198 241 920 417 792 × 2 = 0 + 0.001 556 396 483 840 835 584;
  • 22) 0.001 556 396 483 840 835 584 × 2 = 0 + 0.003 112 792 967 681 671 168;
  • 23) 0.003 112 792 967 681 671 168 × 2 = 0 + 0.006 225 585 935 363 342 336;
  • 24) 0.006 225 585 935 363 342 336 × 2 = 0 + 0.012 451 171 870 726 684 672;
  • 25) 0.012 451 171 870 726 684 672 × 2 = 0 + 0.024 902 343 741 453 369 344;
  • 26) 0.024 902 343 741 453 369 344 × 2 = 0 + 0.049 804 687 482 906 738 688;
  • 27) 0.049 804 687 482 906 738 688 × 2 = 0 + 0.099 609 374 965 813 477 376;
  • 28) 0.099 609 374 965 813 477 376 × 2 = 0 + 0.199 218 749 931 626 954 752;
  • 29) 0.199 218 749 931 626 954 752 × 2 = 0 + 0.398 437 499 863 253 909 504;
  • 30) 0.398 437 499 863 253 909 504 × 2 = 0 + 0.796 874 999 726 507 819 008;
  • 31) 0.796 874 999 726 507 819 008 × 2 = 1 + 0.593 749 999 453 015 638 016;
  • 32) 0.593 749 999 453 015 638 016 × 2 = 1 + 0.187 499 998 906 031 276 032;
  • 33) 0.187 499 998 906 031 276 032 × 2 = 0 + 0.374 999 997 812 062 552 064;
  • 34) 0.374 999 997 812 062 552 064 × 2 = 0 + 0.749 999 995 624 125 104 128;
  • 35) 0.749 999 995 624 125 104 128 × 2 = 1 + 0.499 999 991 248 250 208 256;
  • 36) 0.499 999 991 248 250 208 256 × 2 = 0 + 0.999 999 982 496 500 416 512;
  • 37) 0.999 999 982 496 500 416 512 × 2 = 1 + 0.999 999 964 993 000 833 024;
  • 38) 0.999 999 964 993 000 833 024 × 2 = 1 + 0.999 999 929 986 001 666 048;
  • 39) 0.999 999 929 986 001 666 048 × 2 = 1 + 0.999 999 859 972 003 332 096;
  • 40) 0.999 999 859 972 003 332 096 × 2 = 1 + 0.999 999 719 944 006 664 192;
  • 41) 0.999 999 719 944 006 664 192 × 2 = 1 + 0.999 999 439 888 013 328 384;
  • 42) 0.999 999 439 888 013 328 384 × 2 = 1 + 0.999 998 879 776 026 656 768;
  • 43) 0.999 998 879 776 026 656 768 × 2 = 1 + 0.999 997 759 552 053 313 536;
  • 44) 0.999 997 759 552 053 313 536 × 2 = 1 + 0.999 995 519 104 106 627 072;
  • 45) 0.999 995 519 104 106 627 072 × 2 = 1 + 0.999 991 038 208 213 254 144;
  • 46) 0.999 991 038 208 213 254 144 × 2 = 1 + 0.999 982 076 416 426 508 288;
  • 47) 0.999 982 076 416 426 508 288 × 2 = 1 + 0.999 964 152 832 853 016 576;
  • 48) 0.999 964 152 832 853 016 576 × 2 = 1 + 0.999 928 305 665 706 033 152;
  • 49) 0.999 928 305 665 706 033 152 × 2 = 1 + 0.999 856 611 331 412 066 304;
  • 50) 0.999 856 611 331 412 066 304 × 2 = 1 + 0.999 713 222 662 824 132 608;
  • 51) 0.999 713 222 662 824 132 608 × 2 = 1 + 0.999 426 445 325 648 265 216;
  • 52) 0.999 426 445 325 648 265 216 × 2 = 1 + 0.998 852 890 651 296 530 432;
  • 53) 0.998 852 890 651 296 530 432 × 2 = 1 + 0.997 705 781 302 593 060 864;
  • 54) 0.997 705 781 302 593 060 864 × 2 = 1 + 0.995 411 562 605 186 121 728;

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:


0.000 000 000 742 147 676 392(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 392(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:


0.000 000 000 742 147 676 392(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 -0.000 000 000 742 147 676 392 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