-0.000 000 000 742 147 670 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 670 3(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
-0.000 000 000 742 147 670 3(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:

|-0.000 000 000 742 147 670 3| = 0.000 000 000 742 147 670 3


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: 0.000 000 000 742 147 670 3.

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) 0.000 000 000 742 147 670 3 × 2 = 0 + 0.000 000 001 484 295 340 6;
  • 2) 0.000 000 001 484 295 340 6 × 2 = 0 + 0.000 000 002 968 590 681 2;
  • 3) 0.000 000 002 968 590 681 2 × 2 = 0 + 0.000 000 005 937 181 362 4;
  • 4) 0.000 000 005 937 181 362 4 × 2 = 0 + 0.000 000 011 874 362 724 8;
  • 5) 0.000 000 011 874 362 724 8 × 2 = 0 + 0.000 000 023 748 725 449 6;
  • 6) 0.000 000 023 748 725 449 6 × 2 = 0 + 0.000 000 047 497 450 899 2;
  • 7) 0.000 000 047 497 450 899 2 × 2 = 0 + 0.000 000 094 994 901 798 4;
  • 8) 0.000 000 094 994 901 798 4 × 2 = 0 + 0.000 000 189 989 803 596 8;
  • 9) 0.000 000 189 989 803 596 8 × 2 = 0 + 0.000 000 379 979 607 193 6;
  • 10) 0.000 000 379 979 607 193 6 × 2 = 0 + 0.000 000 759 959 214 387 2;
  • 11) 0.000 000 759 959 214 387 2 × 2 = 0 + 0.000 001 519 918 428 774 4;
  • 12) 0.000 001 519 918 428 774 4 × 2 = 0 + 0.000 003 039 836 857 548 8;
  • 13) 0.000 003 039 836 857 548 8 × 2 = 0 + 0.000 006 079 673 715 097 6;
  • 14) 0.000 006 079 673 715 097 6 × 2 = 0 + 0.000 012 159 347 430 195 2;
  • 15) 0.000 012 159 347 430 195 2 × 2 = 0 + 0.000 024 318 694 860 390 4;
  • 16) 0.000 024 318 694 860 390 4 × 2 = 0 + 0.000 048 637 389 720 780 8;
  • 17) 0.000 048 637 389 720 780 8 × 2 = 0 + 0.000 097 274 779 441 561 6;
  • 18) 0.000 097 274 779 441 561 6 × 2 = 0 + 0.000 194 549 558 883 123 2;
  • 19) 0.000 194 549 558 883 123 2 × 2 = 0 + 0.000 389 099 117 766 246 4;
  • 20) 0.000 389 099 117 766 246 4 × 2 = 0 + 0.000 778 198 235 532 492 8;
  • 21) 0.000 778 198 235 532 492 8 × 2 = 0 + 0.001 556 396 471 064 985 6;
  • 22) 0.001 556 396 471 064 985 6 × 2 = 0 + 0.003 112 792 942 129 971 2;
  • 23) 0.003 112 792 942 129 971 2 × 2 = 0 + 0.006 225 585 884 259 942 4;
  • 24) 0.006 225 585 884 259 942 4 × 2 = 0 + 0.012 451 171 768 519 884 8;
  • 25) 0.012 451 171 768 519 884 8 × 2 = 0 + 0.024 902 343 537 039 769 6;
  • 26) 0.024 902 343 537 039 769 6 × 2 = 0 + 0.049 804 687 074 079 539 2;
  • 27) 0.049 804 687 074 079 539 2 × 2 = 0 + 0.099 609 374 148 159 078 4;
  • 28) 0.099 609 374 148 159 078 4 × 2 = 0 + 0.199 218 748 296 318 156 8;
  • 29) 0.199 218 748 296 318 156 8 × 2 = 0 + 0.398 437 496 592 636 313 6;
  • 30) 0.398 437 496 592 636 313 6 × 2 = 0 + 0.796 874 993 185 272 627 2;
  • 31) 0.796 874 993 185 272 627 2 × 2 = 1 + 0.593 749 986 370 545 254 4;
  • 32) 0.593 749 986 370 545 254 4 × 2 = 1 + 0.187 499 972 741 090 508 8;
  • 33) 0.187 499 972 741 090 508 8 × 2 = 0 + 0.374 999 945 482 181 017 6;
  • 34) 0.374 999 945 482 181 017 6 × 2 = 0 + 0.749 999 890 964 362 035 2;
  • 35) 0.749 999 890 964 362 035 2 × 2 = 1 + 0.499 999 781 928 724 070 4;
  • 36) 0.499 999 781 928 724 070 4 × 2 = 0 + 0.999 999 563 857 448 140 8;
  • 37) 0.999 999 563 857 448 140 8 × 2 = 1 + 0.999 999 127 714 896 281 6;
  • 38) 0.999 999 127 714 896 281 6 × 2 = 1 + 0.999 998 255 429 792 563 2;
  • 39) 0.999 998 255 429 792 563 2 × 2 = 1 + 0.999 996 510 859 585 126 4;
  • 40) 0.999 996 510 859 585 126 4 × 2 = 1 + 0.999 993 021 719 170 252 8;
  • 41) 0.999 993 021 719 170 252 8 × 2 = 1 + 0.999 986 043 438 340 505 6;
  • 42) 0.999 986 043 438 340 505 6 × 2 = 1 + 0.999 972 086 876 681 011 2;
  • 43) 0.999 972 086 876 681 011 2 × 2 = 1 + 0.999 944 173 753 362 022 4;
  • 44) 0.999 944 173 753 362 022 4 × 2 = 1 + 0.999 888 347 506 724 044 8;
  • 45) 0.999 888 347 506 724 044 8 × 2 = 1 + 0.999 776 695 013 448 089 6;
  • 46) 0.999 776 695 013 448 089 6 × 2 = 1 + 0.999 553 390 026 896 179 2;
  • 47) 0.999 553 390 026 896 179 2 × 2 = 1 + 0.999 106 780 053 792 358 4;
  • 48) 0.999 106 780 053 792 358 4 × 2 = 1 + 0.998 213 560 107 584 716 8;
  • 49) 0.998 213 560 107 584 716 8 × 2 = 1 + 0.996 427 120 215 169 433 6;
  • 50) 0.996 427 120 215 169 433 6 × 2 = 1 + 0.992 854 240 430 338 867 2;
  • 51) 0.992 854 240 430 338 867 2 × 2 = 1 + 0.985 708 480 860 677 734 4;
  • 52) 0.985 708 480 860 677 734 4 × 2 = 1 + 0.971 416 961 721 355 468 8;
  • 53) 0.971 416 961 721 355 468 8 × 2 = 1 + 0.942 833 923 442 710 937 6;
  • 54) 0.942 833 923 442 710 937 6 × 2 = 1 + 0.885 667 846 885 421 875 2;

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:


0.000 000 000 742 147 670 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 670 3(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:


0.000 000 000 742 147 670 3(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 -0.000 000 000 742 147 670 3 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