-0.000 000 000 742 147 674 48 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 48(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
-0.000 000 000 742 147 674 48(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:

|-0.000 000 000 742 147 674 48| = 0.000 000 000 742 147 674 48


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: 0.000 000 000 742 147 674 48.

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) 0.000 000 000 742 147 674 48 × 2 = 0 + 0.000 000 001 484 295 348 96;
  • 2) 0.000 000 001 484 295 348 96 × 2 = 0 + 0.000 000 002 968 590 697 92;
  • 3) 0.000 000 002 968 590 697 92 × 2 = 0 + 0.000 000 005 937 181 395 84;
  • 4) 0.000 000 005 937 181 395 84 × 2 = 0 + 0.000 000 011 874 362 791 68;
  • 5) 0.000 000 011 874 362 791 68 × 2 = 0 + 0.000 000 023 748 725 583 36;
  • 6) 0.000 000 023 748 725 583 36 × 2 = 0 + 0.000 000 047 497 451 166 72;
  • 7) 0.000 000 047 497 451 166 72 × 2 = 0 + 0.000 000 094 994 902 333 44;
  • 8) 0.000 000 094 994 902 333 44 × 2 = 0 + 0.000 000 189 989 804 666 88;
  • 9) 0.000 000 189 989 804 666 88 × 2 = 0 + 0.000 000 379 979 609 333 76;
  • 10) 0.000 000 379 979 609 333 76 × 2 = 0 + 0.000 000 759 959 218 667 52;
  • 11) 0.000 000 759 959 218 667 52 × 2 = 0 + 0.000 001 519 918 437 335 04;
  • 12) 0.000 001 519 918 437 335 04 × 2 = 0 + 0.000 003 039 836 874 670 08;
  • 13) 0.000 003 039 836 874 670 08 × 2 = 0 + 0.000 006 079 673 749 340 16;
  • 14) 0.000 006 079 673 749 340 16 × 2 = 0 + 0.000 012 159 347 498 680 32;
  • 15) 0.000 012 159 347 498 680 32 × 2 = 0 + 0.000 024 318 694 997 360 64;
  • 16) 0.000 024 318 694 997 360 64 × 2 = 0 + 0.000 048 637 389 994 721 28;
  • 17) 0.000 048 637 389 994 721 28 × 2 = 0 + 0.000 097 274 779 989 442 56;
  • 18) 0.000 097 274 779 989 442 56 × 2 = 0 + 0.000 194 549 559 978 885 12;
  • 19) 0.000 194 549 559 978 885 12 × 2 = 0 + 0.000 389 099 119 957 770 24;
  • 20) 0.000 389 099 119 957 770 24 × 2 = 0 + 0.000 778 198 239 915 540 48;
  • 21) 0.000 778 198 239 915 540 48 × 2 = 0 + 0.001 556 396 479 831 080 96;
  • 22) 0.001 556 396 479 831 080 96 × 2 = 0 + 0.003 112 792 959 662 161 92;
  • 23) 0.003 112 792 959 662 161 92 × 2 = 0 + 0.006 225 585 919 324 323 84;
  • 24) 0.006 225 585 919 324 323 84 × 2 = 0 + 0.012 451 171 838 648 647 68;
  • 25) 0.012 451 171 838 648 647 68 × 2 = 0 + 0.024 902 343 677 297 295 36;
  • 26) 0.024 902 343 677 297 295 36 × 2 = 0 + 0.049 804 687 354 594 590 72;
  • 27) 0.049 804 687 354 594 590 72 × 2 = 0 + 0.099 609 374 709 189 181 44;
  • 28) 0.099 609 374 709 189 181 44 × 2 = 0 + 0.199 218 749 418 378 362 88;
  • 29) 0.199 218 749 418 378 362 88 × 2 = 0 + 0.398 437 498 836 756 725 76;
  • 30) 0.398 437 498 836 756 725 76 × 2 = 0 + 0.796 874 997 673 513 451 52;
  • 31) 0.796 874 997 673 513 451 52 × 2 = 1 + 0.593 749 995 347 026 903 04;
  • 32) 0.593 749 995 347 026 903 04 × 2 = 1 + 0.187 499 990 694 053 806 08;
  • 33) 0.187 499 990 694 053 806 08 × 2 = 0 + 0.374 999 981 388 107 612 16;
  • 34) 0.374 999 981 388 107 612 16 × 2 = 0 + 0.749 999 962 776 215 224 32;
  • 35) 0.749 999 962 776 215 224 32 × 2 = 1 + 0.499 999 925 552 430 448 64;
  • 36) 0.499 999 925 552 430 448 64 × 2 = 0 + 0.999 999 851 104 860 897 28;
  • 37) 0.999 999 851 104 860 897 28 × 2 = 1 + 0.999 999 702 209 721 794 56;
  • 38) 0.999 999 702 209 721 794 56 × 2 = 1 + 0.999 999 404 419 443 589 12;
  • 39) 0.999 999 404 419 443 589 12 × 2 = 1 + 0.999 998 808 838 887 178 24;
  • 40) 0.999 998 808 838 887 178 24 × 2 = 1 + 0.999 997 617 677 774 356 48;
  • 41) 0.999 997 617 677 774 356 48 × 2 = 1 + 0.999 995 235 355 548 712 96;
  • 42) 0.999 995 235 355 548 712 96 × 2 = 1 + 0.999 990 470 711 097 425 92;
  • 43) 0.999 990 470 711 097 425 92 × 2 = 1 + 0.999 980 941 422 194 851 84;
  • 44) 0.999 980 941 422 194 851 84 × 2 = 1 + 0.999 961 882 844 389 703 68;
  • 45) 0.999 961 882 844 389 703 68 × 2 = 1 + 0.999 923 765 688 779 407 36;
  • 46) 0.999 923 765 688 779 407 36 × 2 = 1 + 0.999 847 531 377 558 814 72;
  • 47) 0.999 847 531 377 558 814 72 × 2 = 1 + 0.999 695 062 755 117 629 44;
  • 48) 0.999 695 062 755 117 629 44 × 2 = 1 + 0.999 390 125 510 235 258 88;
  • 49) 0.999 390 125 510 235 258 88 × 2 = 1 + 0.998 780 251 020 470 517 76;
  • 50) 0.998 780 251 020 470 517 76 × 2 = 1 + 0.997 560 502 040 941 035 52;
  • 51) 0.997 560 502 040 941 035 52 × 2 = 1 + 0.995 121 004 081 882 071 04;
  • 52) 0.995 121 004 081 882 071 04 × 2 = 1 + 0.990 242 008 163 764 142 08;
  • 53) 0.990 242 008 163 764 142 08 × 2 = 1 + 0.980 484 016 327 528 284 16;
  • 54) 0.980 484 016 327 528 284 16 × 2 = 1 + 0.960 968 032 655 056 568 32;

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:


0.000 000 000 742 147 674 48(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674 48(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:


0.000 000 000 742 147 674 48(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 -0.000 000 000 742 147 674 48 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