-0.000 000 000 742 016 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 016(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
-0.000 000 000 742 016(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:

|-0.000 000 000 742 016| = 0.000 000 000 742 016


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: 0.000 000 000 742 016.

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) 0.000 000 000 742 016 × 2 = 0 + 0.000 000 001 484 032;
  • 2) 0.000 000 001 484 032 × 2 = 0 + 0.000 000 002 968 064;
  • 3) 0.000 000 002 968 064 × 2 = 0 + 0.000 000 005 936 128;
  • 4) 0.000 000 005 936 128 × 2 = 0 + 0.000 000 011 872 256;
  • 5) 0.000 000 011 872 256 × 2 = 0 + 0.000 000 023 744 512;
  • 6) 0.000 000 023 744 512 × 2 = 0 + 0.000 000 047 489 024;
  • 7) 0.000 000 047 489 024 × 2 = 0 + 0.000 000 094 978 048;
  • 8) 0.000 000 094 978 048 × 2 = 0 + 0.000 000 189 956 096;
  • 9) 0.000 000 189 956 096 × 2 = 0 + 0.000 000 379 912 192;
  • 10) 0.000 000 379 912 192 × 2 = 0 + 0.000 000 759 824 384;
  • 11) 0.000 000 759 824 384 × 2 = 0 + 0.000 001 519 648 768;
  • 12) 0.000 001 519 648 768 × 2 = 0 + 0.000 003 039 297 536;
  • 13) 0.000 003 039 297 536 × 2 = 0 + 0.000 006 078 595 072;
  • 14) 0.000 006 078 595 072 × 2 = 0 + 0.000 012 157 190 144;
  • 15) 0.000 012 157 190 144 × 2 = 0 + 0.000 024 314 380 288;
  • 16) 0.000 024 314 380 288 × 2 = 0 + 0.000 048 628 760 576;
  • 17) 0.000 048 628 760 576 × 2 = 0 + 0.000 097 257 521 152;
  • 18) 0.000 097 257 521 152 × 2 = 0 + 0.000 194 515 042 304;
  • 19) 0.000 194 515 042 304 × 2 = 0 + 0.000 389 030 084 608;
  • 20) 0.000 389 030 084 608 × 2 = 0 + 0.000 778 060 169 216;
  • 21) 0.000 778 060 169 216 × 2 = 0 + 0.001 556 120 338 432;
  • 22) 0.001 556 120 338 432 × 2 = 0 + 0.003 112 240 676 864;
  • 23) 0.003 112 240 676 864 × 2 = 0 + 0.006 224 481 353 728;
  • 24) 0.006 224 481 353 728 × 2 = 0 + 0.012 448 962 707 456;
  • 25) 0.012 448 962 707 456 × 2 = 0 + 0.024 897 925 414 912;
  • 26) 0.024 897 925 414 912 × 2 = 0 + 0.049 795 850 829 824;
  • 27) 0.049 795 850 829 824 × 2 = 0 + 0.099 591 701 659 648;
  • 28) 0.099 591 701 659 648 × 2 = 0 + 0.199 183 403 319 296;
  • 29) 0.199 183 403 319 296 × 2 = 0 + 0.398 366 806 638 592;
  • 30) 0.398 366 806 638 592 × 2 = 0 + 0.796 733 613 277 184;
  • 31) 0.796 733 613 277 184 × 2 = 1 + 0.593 467 226 554 368;
  • 32) 0.593 467 226 554 368 × 2 = 1 + 0.186 934 453 108 736;
  • 33) 0.186 934 453 108 736 × 2 = 0 + 0.373 868 906 217 472;
  • 34) 0.373 868 906 217 472 × 2 = 0 + 0.747 737 812 434 944;
  • 35) 0.747 737 812 434 944 × 2 = 1 + 0.495 475 624 869 888;
  • 36) 0.495 475 624 869 888 × 2 = 0 + 0.990 951 249 739 776;
  • 37) 0.990 951 249 739 776 × 2 = 1 + 0.981 902 499 479 552;
  • 38) 0.981 902 499 479 552 × 2 = 1 + 0.963 804 998 959 104;
  • 39) 0.963 804 998 959 104 × 2 = 1 + 0.927 609 997 918 208;
  • 40) 0.927 609 997 918 208 × 2 = 1 + 0.855 219 995 836 416;
  • 41) 0.855 219 995 836 416 × 2 = 1 + 0.710 439 991 672 832;
  • 42) 0.710 439 991 672 832 × 2 = 1 + 0.420 879 983 345 664;
  • 43) 0.420 879 983 345 664 × 2 = 0 + 0.841 759 966 691 328;
  • 44) 0.841 759 966 691 328 × 2 = 1 + 0.683 519 933 382 656;
  • 45) 0.683 519 933 382 656 × 2 = 1 + 0.367 039 866 765 312;
  • 46) 0.367 039 866 765 312 × 2 = 0 + 0.734 079 733 530 624;
  • 47) 0.734 079 733 530 624 × 2 = 1 + 0.468 159 467 061 248;
  • 48) 0.468 159 467 061 248 × 2 = 0 + 0.936 318 934 122 496;
  • 49) 0.936 318 934 122 496 × 2 = 1 + 0.872 637 868 244 992;
  • 50) 0.872 637 868 244 992 × 2 = 1 + 0.745 275 736 489 984;
  • 51) 0.745 275 736 489 984 × 2 = 1 + 0.490 551 472 979 968;
  • 52) 0.490 551 472 979 968 × 2 = 0 + 0.981 102 945 959 936;
  • 53) 0.981 102 945 959 936 × 2 = 1 + 0.962 205 891 919 872;
  • 54) 0.962 205 891 919 872 × 2 = 1 + 0.924 411 783 839 744;

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:


0.000 000 000 742 016(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1010 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 016(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1010 1110 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:


0.000 000 000 742 016(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1010 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 1010 1110 11(2) × 20 =

1.1001 0111 1110 1101 0111 011(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 1110 1101 0111 011


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 0110 1011 1011 =

100 1011 1111 0110 1011 1011


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 0110 1011 1011


Decimal number -0.000 000 000 742 016 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0110 1011 1011

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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