-0.000 000 000 742 147 676 620 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 620 6(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
-0.000 000 000 742 147 676 620 6(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:

|-0.000 000 000 742 147 676 620 6| = 0.000 000 000 742 147 676 620 6


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: 0.000 000 000 742 147 676 620 6.

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) 0.000 000 000 742 147 676 620 6 × 2 = 0 + 0.000 000 001 484 295 353 241 2;
  • 2) 0.000 000 001 484 295 353 241 2 × 2 = 0 + 0.000 000 002 968 590 706 482 4;
  • 3) 0.000 000 002 968 590 706 482 4 × 2 = 0 + 0.000 000 005 937 181 412 964 8;
  • 4) 0.000 000 005 937 181 412 964 8 × 2 = 0 + 0.000 000 011 874 362 825 929 6;
  • 5) 0.000 000 011 874 362 825 929 6 × 2 = 0 + 0.000 000 023 748 725 651 859 2;
  • 6) 0.000 000 023 748 725 651 859 2 × 2 = 0 + 0.000 000 047 497 451 303 718 4;
  • 7) 0.000 000 047 497 451 303 718 4 × 2 = 0 + 0.000 000 094 994 902 607 436 8;
  • 8) 0.000 000 094 994 902 607 436 8 × 2 = 0 + 0.000 000 189 989 805 214 873 6;
  • 9) 0.000 000 189 989 805 214 873 6 × 2 = 0 + 0.000 000 379 979 610 429 747 2;
  • 10) 0.000 000 379 979 610 429 747 2 × 2 = 0 + 0.000 000 759 959 220 859 494 4;
  • 11) 0.000 000 759 959 220 859 494 4 × 2 = 0 + 0.000 001 519 918 441 718 988 8;
  • 12) 0.000 001 519 918 441 718 988 8 × 2 = 0 + 0.000 003 039 836 883 437 977 6;
  • 13) 0.000 003 039 836 883 437 977 6 × 2 = 0 + 0.000 006 079 673 766 875 955 2;
  • 14) 0.000 006 079 673 766 875 955 2 × 2 = 0 + 0.000 012 159 347 533 751 910 4;
  • 15) 0.000 012 159 347 533 751 910 4 × 2 = 0 + 0.000 024 318 695 067 503 820 8;
  • 16) 0.000 024 318 695 067 503 820 8 × 2 = 0 + 0.000 048 637 390 135 007 641 6;
  • 17) 0.000 048 637 390 135 007 641 6 × 2 = 0 + 0.000 097 274 780 270 015 283 2;
  • 18) 0.000 097 274 780 270 015 283 2 × 2 = 0 + 0.000 194 549 560 540 030 566 4;
  • 19) 0.000 194 549 560 540 030 566 4 × 2 = 0 + 0.000 389 099 121 080 061 132 8;
  • 20) 0.000 389 099 121 080 061 132 8 × 2 = 0 + 0.000 778 198 242 160 122 265 6;
  • 21) 0.000 778 198 242 160 122 265 6 × 2 = 0 + 0.001 556 396 484 320 244 531 2;
  • 22) 0.001 556 396 484 320 244 531 2 × 2 = 0 + 0.003 112 792 968 640 489 062 4;
  • 23) 0.003 112 792 968 640 489 062 4 × 2 = 0 + 0.006 225 585 937 280 978 124 8;
  • 24) 0.006 225 585 937 280 978 124 8 × 2 = 0 + 0.012 451 171 874 561 956 249 6;
  • 25) 0.012 451 171 874 561 956 249 6 × 2 = 0 + 0.024 902 343 749 123 912 499 2;
  • 26) 0.024 902 343 749 123 912 499 2 × 2 = 0 + 0.049 804 687 498 247 824 998 4;
  • 27) 0.049 804 687 498 247 824 998 4 × 2 = 0 + 0.099 609 374 996 495 649 996 8;
  • 28) 0.099 609 374 996 495 649 996 8 × 2 = 0 + 0.199 218 749 992 991 299 993 6;
  • 29) 0.199 218 749 992 991 299 993 6 × 2 = 0 + 0.398 437 499 985 982 599 987 2;
  • 30) 0.398 437 499 985 982 599 987 2 × 2 = 0 + 0.796 874 999 971 965 199 974 4;
  • 31) 0.796 874 999 971 965 199 974 4 × 2 = 1 + 0.593 749 999 943 930 399 948 8;
  • 32) 0.593 749 999 943 930 399 948 8 × 2 = 1 + 0.187 499 999 887 860 799 897 6;
  • 33) 0.187 499 999 887 860 799 897 6 × 2 = 0 + 0.374 999 999 775 721 599 795 2;
  • 34) 0.374 999 999 775 721 599 795 2 × 2 = 0 + 0.749 999 999 551 443 199 590 4;
  • 35) 0.749 999 999 551 443 199 590 4 × 2 = 1 + 0.499 999 999 102 886 399 180 8;
  • 36) 0.499 999 999 102 886 399 180 8 × 2 = 0 + 0.999 999 998 205 772 798 361 6;
  • 37) 0.999 999 998 205 772 798 361 6 × 2 = 1 + 0.999 999 996 411 545 596 723 2;
  • 38) 0.999 999 996 411 545 596 723 2 × 2 = 1 + 0.999 999 992 823 091 193 446 4;
  • 39) 0.999 999 992 823 091 193 446 4 × 2 = 1 + 0.999 999 985 646 182 386 892 8;
  • 40) 0.999 999 985 646 182 386 892 8 × 2 = 1 + 0.999 999 971 292 364 773 785 6;
  • 41) 0.999 999 971 292 364 773 785 6 × 2 = 1 + 0.999 999 942 584 729 547 571 2;
  • 42) 0.999 999 942 584 729 547 571 2 × 2 = 1 + 0.999 999 885 169 459 095 142 4;
  • 43) 0.999 999 885 169 459 095 142 4 × 2 = 1 + 0.999 999 770 338 918 190 284 8;
  • 44) 0.999 999 770 338 918 190 284 8 × 2 = 1 + 0.999 999 540 677 836 380 569 6;
  • 45) 0.999 999 540 677 836 380 569 6 × 2 = 1 + 0.999 999 081 355 672 761 139 2;
  • 46) 0.999 999 081 355 672 761 139 2 × 2 = 1 + 0.999 998 162 711 345 522 278 4;
  • 47) 0.999 998 162 711 345 522 278 4 × 2 = 1 + 0.999 996 325 422 691 044 556 8;
  • 48) 0.999 996 325 422 691 044 556 8 × 2 = 1 + 0.999 992 650 845 382 089 113 6;
  • 49) 0.999 992 650 845 382 089 113 6 × 2 = 1 + 0.999 985 301 690 764 178 227 2;
  • 50) 0.999 985 301 690 764 178 227 2 × 2 = 1 + 0.999 970 603 381 528 356 454 4;
  • 51) 0.999 970 603 381 528 356 454 4 × 2 = 1 + 0.999 941 206 763 056 712 908 8;
  • 52) 0.999 941 206 763 056 712 908 8 × 2 = 1 + 0.999 882 413 526 113 425 817 6;
  • 53) 0.999 882 413 526 113 425 817 6 × 2 = 1 + 0.999 764 827 052 226 851 635 2;
  • 54) 0.999 764 827 052 226 851 635 2 × 2 = 1 + 0.999 529 654 104 453 703 270 4;

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:


0.000 000 000 742 147 676 620 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 620 6(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:


0.000 000 000 742 147 676 620 6(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 -0.000 000 000 742 147 676 620 6 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