-0.000 000 000 742 147 676 646 696 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 696 2(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
-0.000 000 000 742 147 676 646 696 2(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:

|-0.000 000 000 742 147 676 646 696 2| = 0.000 000 000 742 147 676 646 696 2


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: 0.000 000 000 742 147 676 646 696 2.

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) 0.000 000 000 742 147 676 646 696 2 × 2 = 0 + 0.000 000 001 484 295 353 293 392 4;
  • 2) 0.000 000 001 484 295 353 293 392 4 × 2 = 0 + 0.000 000 002 968 590 706 586 784 8;
  • 3) 0.000 000 002 968 590 706 586 784 8 × 2 = 0 + 0.000 000 005 937 181 413 173 569 6;
  • 4) 0.000 000 005 937 181 413 173 569 6 × 2 = 0 + 0.000 000 011 874 362 826 347 139 2;
  • 5) 0.000 000 011 874 362 826 347 139 2 × 2 = 0 + 0.000 000 023 748 725 652 694 278 4;
  • 6) 0.000 000 023 748 725 652 694 278 4 × 2 = 0 + 0.000 000 047 497 451 305 388 556 8;
  • 7) 0.000 000 047 497 451 305 388 556 8 × 2 = 0 + 0.000 000 094 994 902 610 777 113 6;
  • 8) 0.000 000 094 994 902 610 777 113 6 × 2 = 0 + 0.000 000 189 989 805 221 554 227 2;
  • 9) 0.000 000 189 989 805 221 554 227 2 × 2 = 0 + 0.000 000 379 979 610 443 108 454 4;
  • 10) 0.000 000 379 979 610 443 108 454 4 × 2 = 0 + 0.000 000 759 959 220 886 216 908 8;
  • 11) 0.000 000 759 959 220 886 216 908 8 × 2 = 0 + 0.000 001 519 918 441 772 433 817 6;
  • 12) 0.000 001 519 918 441 772 433 817 6 × 2 = 0 + 0.000 003 039 836 883 544 867 635 2;
  • 13) 0.000 003 039 836 883 544 867 635 2 × 2 = 0 + 0.000 006 079 673 767 089 735 270 4;
  • 14) 0.000 006 079 673 767 089 735 270 4 × 2 = 0 + 0.000 012 159 347 534 179 470 540 8;
  • 15) 0.000 012 159 347 534 179 470 540 8 × 2 = 0 + 0.000 024 318 695 068 358 941 081 6;
  • 16) 0.000 024 318 695 068 358 941 081 6 × 2 = 0 + 0.000 048 637 390 136 717 882 163 2;
  • 17) 0.000 048 637 390 136 717 882 163 2 × 2 = 0 + 0.000 097 274 780 273 435 764 326 4;
  • 18) 0.000 097 274 780 273 435 764 326 4 × 2 = 0 + 0.000 194 549 560 546 871 528 652 8;
  • 19) 0.000 194 549 560 546 871 528 652 8 × 2 = 0 + 0.000 389 099 121 093 743 057 305 6;
  • 20) 0.000 389 099 121 093 743 057 305 6 × 2 = 0 + 0.000 778 198 242 187 486 114 611 2;
  • 21) 0.000 778 198 242 187 486 114 611 2 × 2 = 0 + 0.001 556 396 484 374 972 229 222 4;
  • 22) 0.001 556 396 484 374 972 229 222 4 × 2 = 0 + 0.003 112 792 968 749 944 458 444 8;
  • 23) 0.003 112 792 968 749 944 458 444 8 × 2 = 0 + 0.006 225 585 937 499 888 916 889 6;
  • 24) 0.006 225 585 937 499 888 916 889 6 × 2 = 0 + 0.012 451 171 874 999 777 833 779 2;
  • 25) 0.012 451 171 874 999 777 833 779 2 × 2 = 0 + 0.024 902 343 749 999 555 667 558 4;
  • 26) 0.024 902 343 749 999 555 667 558 4 × 2 = 0 + 0.049 804 687 499 999 111 335 116 8;
  • 27) 0.049 804 687 499 999 111 335 116 8 × 2 = 0 + 0.099 609 374 999 998 222 670 233 6;
  • 28) 0.099 609 374 999 998 222 670 233 6 × 2 = 0 + 0.199 218 749 999 996 445 340 467 2;
  • 29) 0.199 218 749 999 996 445 340 467 2 × 2 = 0 + 0.398 437 499 999 992 890 680 934 4;
  • 30) 0.398 437 499 999 992 890 680 934 4 × 2 = 0 + 0.796 874 999 999 985 781 361 868 8;
  • 31) 0.796 874 999 999 985 781 361 868 8 × 2 = 1 + 0.593 749 999 999 971 562 723 737 6;
  • 32) 0.593 749 999 999 971 562 723 737 6 × 2 = 1 + 0.187 499 999 999 943 125 447 475 2;
  • 33) 0.187 499 999 999 943 125 447 475 2 × 2 = 0 + 0.374 999 999 999 886 250 894 950 4;
  • 34) 0.374 999 999 999 886 250 894 950 4 × 2 = 0 + 0.749 999 999 999 772 501 789 900 8;
  • 35) 0.749 999 999 999 772 501 789 900 8 × 2 = 1 + 0.499 999 999 999 545 003 579 801 6;
  • 36) 0.499 999 999 999 545 003 579 801 6 × 2 = 0 + 0.999 999 999 999 090 007 159 603 2;
  • 37) 0.999 999 999 999 090 007 159 603 2 × 2 = 1 + 0.999 999 999 998 180 014 319 206 4;
  • 38) 0.999 999 999 998 180 014 319 206 4 × 2 = 1 + 0.999 999 999 996 360 028 638 412 8;
  • 39) 0.999 999 999 996 360 028 638 412 8 × 2 = 1 + 0.999 999 999 992 720 057 276 825 6;
  • 40) 0.999 999 999 992 720 057 276 825 6 × 2 = 1 + 0.999 999 999 985 440 114 553 651 2;
  • 41) 0.999 999 999 985 440 114 553 651 2 × 2 = 1 + 0.999 999 999 970 880 229 107 302 4;
  • 42) 0.999 999 999 970 880 229 107 302 4 × 2 = 1 + 0.999 999 999 941 760 458 214 604 8;
  • 43) 0.999 999 999 941 760 458 214 604 8 × 2 = 1 + 0.999 999 999 883 520 916 429 209 6;
  • 44) 0.999 999 999 883 520 916 429 209 6 × 2 = 1 + 0.999 999 999 767 041 832 858 419 2;
  • 45) 0.999 999 999 767 041 832 858 419 2 × 2 = 1 + 0.999 999 999 534 083 665 716 838 4;
  • 46) 0.999 999 999 534 083 665 716 838 4 × 2 = 1 + 0.999 999 999 068 167 331 433 676 8;
  • 47) 0.999 999 999 068 167 331 433 676 8 × 2 = 1 + 0.999 999 998 136 334 662 867 353 6;
  • 48) 0.999 999 998 136 334 662 867 353 6 × 2 = 1 + 0.999 999 996 272 669 325 734 707 2;
  • 49) 0.999 999 996 272 669 325 734 707 2 × 2 = 1 + 0.999 999 992 545 338 651 469 414 4;
  • 50) 0.999 999 992 545 338 651 469 414 4 × 2 = 1 + 0.999 999 985 090 677 302 938 828 8;
  • 51) 0.999 999 985 090 677 302 938 828 8 × 2 = 1 + 0.999 999 970 181 354 605 877 657 6;
  • 52) 0.999 999 970 181 354 605 877 657 6 × 2 = 1 + 0.999 999 940 362 709 211 755 315 2;
  • 53) 0.999 999 940 362 709 211 755 315 2 × 2 = 1 + 0.999 999 880 725 418 423 510 630 4;
  • 54) 0.999 999 880 725 418 423 510 630 4 × 2 = 1 + 0.999 999 761 450 836 847 021 260 8;

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:


0.000 000 000 742 147 676 646 696 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 696 2(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:


0.000 000 000 742 147 676 646 696 2(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 -0.000 000 000 742 147 676 646 696 2 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