-0.000 000 000 742 147 489 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 489(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
-0.000 000 000 742 147 489(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:

|-0.000 000 000 742 147 489| = 0.000 000 000 742 147 489


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: 0.000 000 000 742 147 489.

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) 0.000 000 000 742 147 489 × 2 = 0 + 0.000 000 001 484 294 978;
  • 2) 0.000 000 001 484 294 978 × 2 = 0 + 0.000 000 002 968 589 956;
  • 3) 0.000 000 002 968 589 956 × 2 = 0 + 0.000 000 005 937 179 912;
  • 4) 0.000 000 005 937 179 912 × 2 = 0 + 0.000 000 011 874 359 824;
  • 5) 0.000 000 011 874 359 824 × 2 = 0 + 0.000 000 023 748 719 648;
  • 6) 0.000 000 023 748 719 648 × 2 = 0 + 0.000 000 047 497 439 296;
  • 7) 0.000 000 047 497 439 296 × 2 = 0 + 0.000 000 094 994 878 592;
  • 8) 0.000 000 094 994 878 592 × 2 = 0 + 0.000 000 189 989 757 184;
  • 9) 0.000 000 189 989 757 184 × 2 = 0 + 0.000 000 379 979 514 368;
  • 10) 0.000 000 379 979 514 368 × 2 = 0 + 0.000 000 759 959 028 736;
  • 11) 0.000 000 759 959 028 736 × 2 = 0 + 0.000 001 519 918 057 472;
  • 12) 0.000 001 519 918 057 472 × 2 = 0 + 0.000 003 039 836 114 944;
  • 13) 0.000 003 039 836 114 944 × 2 = 0 + 0.000 006 079 672 229 888;
  • 14) 0.000 006 079 672 229 888 × 2 = 0 + 0.000 012 159 344 459 776;
  • 15) 0.000 012 159 344 459 776 × 2 = 0 + 0.000 024 318 688 919 552;
  • 16) 0.000 024 318 688 919 552 × 2 = 0 + 0.000 048 637 377 839 104;
  • 17) 0.000 048 637 377 839 104 × 2 = 0 + 0.000 097 274 755 678 208;
  • 18) 0.000 097 274 755 678 208 × 2 = 0 + 0.000 194 549 511 356 416;
  • 19) 0.000 194 549 511 356 416 × 2 = 0 + 0.000 389 099 022 712 832;
  • 20) 0.000 389 099 022 712 832 × 2 = 0 + 0.000 778 198 045 425 664;
  • 21) 0.000 778 198 045 425 664 × 2 = 0 + 0.001 556 396 090 851 328;
  • 22) 0.001 556 396 090 851 328 × 2 = 0 + 0.003 112 792 181 702 656;
  • 23) 0.003 112 792 181 702 656 × 2 = 0 + 0.006 225 584 363 405 312;
  • 24) 0.006 225 584 363 405 312 × 2 = 0 + 0.012 451 168 726 810 624;
  • 25) 0.012 451 168 726 810 624 × 2 = 0 + 0.024 902 337 453 621 248;
  • 26) 0.024 902 337 453 621 248 × 2 = 0 + 0.049 804 674 907 242 496;
  • 27) 0.049 804 674 907 242 496 × 2 = 0 + 0.099 609 349 814 484 992;
  • 28) 0.099 609 349 814 484 992 × 2 = 0 + 0.199 218 699 628 969 984;
  • 29) 0.199 218 699 628 969 984 × 2 = 0 + 0.398 437 399 257 939 968;
  • 30) 0.398 437 399 257 939 968 × 2 = 0 + 0.796 874 798 515 879 936;
  • 31) 0.796 874 798 515 879 936 × 2 = 1 + 0.593 749 597 031 759 872;
  • 32) 0.593 749 597 031 759 872 × 2 = 1 + 0.187 499 194 063 519 744;
  • 33) 0.187 499 194 063 519 744 × 2 = 0 + 0.374 998 388 127 039 488;
  • 34) 0.374 998 388 127 039 488 × 2 = 0 + 0.749 996 776 254 078 976;
  • 35) 0.749 996 776 254 078 976 × 2 = 1 + 0.499 993 552 508 157 952;
  • 36) 0.499 993 552 508 157 952 × 2 = 0 + 0.999 987 105 016 315 904;
  • 37) 0.999 987 105 016 315 904 × 2 = 1 + 0.999 974 210 032 631 808;
  • 38) 0.999 974 210 032 631 808 × 2 = 1 + 0.999 948 420 065 263 616;
  • 39) 0.999 948 420 065 263 616 × 2 = 1 + 0.999 896 840 130 527 232;
  • 40) 0.999 896 840 130 527 232 × 2 = 1 + 0.999 793 680 261 054 464;
  • 41) 0.999 793 680 261 054 464 × 2 = 1 + 0.999 587 360 522 108 928;
  • 42) 0.999 587 360 522 108 928 × 2 = 1 + 0.999 174 721 044 217 856;
  • 43) 0.999 174 721 044 217 856 × 2 = 1 + 0.998 349 442 088 435 712;
  • 44) 0.998 349 442 088 435 712 × 2 = 1 + 0.996 698 884 176 871 424;
  • 45) 0.996 698 884 176 871 424 × 2 = 1 + 0.993 397 768 353 742 848;
  • 46) 0.993 397 768 353 742 848 × 2 = 1 + 0.986 795 536 707 485 696;
  • 47) 0.986 795 536 707 485 696 × 2 = 1 + 0.973 591 073 414 971 392;
  • 48) 0.973 591 073 414 971 392 × 2 = 1 + 0.947 182 146 829 942 784;
  • 49) 0.947 182 146 829 942 784 × 2 = 1 + 0.894 364 293 659 885 568;
  • 50) 0.894 364 293 659 885 568 × 2 = 1 + 0.788 728 587 319 771 136;
  • 51) 0.788 728 587 319 771 136 × 2 = 1 + 0.577 457 174 639 542 272;
  • 52) 0.577 457 174 639 542 272 × 2 = 1 + 0.154 914 349 279 084 544;
  • 53) 0.154 914 349 279 084 544 × 2 = 0 + 0.309 828 698 558 169 088;
  • 54) 0.309 828 698 558 169 088 × 2 = 0 + 0.619 657 397 116 338 176;

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:


0.000 000 000 742 147 489(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 489(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(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:


0.000 000 000 742 147 489(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 100


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 1100 =

100 1011 1111 1111 1111 1100


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 1100


Decimal number -0.000 000 000 742 147 489 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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