-0.000 000 000 742 147 676 636 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 636 7(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
-0.000 000 000 742 147 676 636 7(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:

|-0.000 000 000 742 147 676 636 7| = 0.000 000 000 742 147 676 636 7


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: 0.000 000 000 742 147 676 636 7.

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) 0.000 000 000 742 147 676 636 7 × 2 = 0 + 0.000 000 001 484 295 353 273 4;
  • 2) 0.000 000 001 484 295 353 273 4 × 2 = 0 + 0.000 000 002 968 590 706 546 8;
  • 3) 0.000 000 002 968 590 706 546 8 × 2 = 0 + 0.000 000 005 937 181 413 093 6;
  • 4) 0.000 000 005 937 181 413 093 6 × 2 = 0 + 0.000 000 011 874 362 826 187 2;
  • 5) 0.000 000 011 874 362 826 187 2 × 2 = 0 + 0.000 000 023 748 725 652 374 4;
  • 6) 0.000 000 023 748 725 652 374 4 × 2 = 0 + 0.000 000 047 497 451 304 748 8;
  • 7) 0.000 000 047 497 451 304 748 8 × 2 = 0 + 0.000 000 094 994 902 609 497 6;
  • 8) 0.000 000 094 994 902 609 497 6 × 2 = 0 + 0.000 000 189 989 805 218 995 2;
  • 9) 0.000 000 189 989 805 218 995 2 × 2 = 0 + 0.000 000 379 979 610 437 990 4;
  • 10) 0.000 000 379 979 610 437 990 4 × 2 = 0 + 0.000 000 759 959 220 875 980 8;
  • 11) 0.000 000 759 959 220 875 980 8 × 2 = 0 + 0.000 001 519 918 441 751 961 6;
  • 12) 0.000 001 519 918 441 751 961 6 × 2 = 0 + 0.000 003 039 836 883 503 923 2;
  • 13) 0.000 003 039 836 883 503 923 2 × 2 = 0 + 0.000 006 079 673 767 007 846 4;
  • 14) 0.000 006 079 673 767 007 846 4 × 2 = 0 + 0.000 012 159 347 534 015 692 8;
  • 15) 0.000 012 159 347 534 015 692 8 × 2 = 0 + 0.000 024 318 695 068 031 385 6;
  • 16) 0.000 024 318 695 068 031 385 6 × 2 = 0 + 0.000 048 637 390 136 062 771 2;
  • 17) 0.000 048 637 390 136 062 771 2 × 2 = 0 + 0.000 097 274 780 272 125 542 4;
  • 18) 0.000 097 274 780 272 125 542 4 × 2 = 0 + 0.000 194 549 560 544 251 084 8;
  • 19) 0.000 194 549 560 544 251 084 8 × 2 = 0 + 0.000 389 099 121 088 502 169 6;
  • 20) 0.000 389 099 121 088 502 169 6 × 2 = 0 + 0.000 778 198 242 177 004 339 2;
  • 21) 0.000 778 198 242 177 004 339 2 × 2 = 0 + 0.001 556 396 484 354 008 678 4;
  • 22) 0.001 556 396 484 354 008 678 4 × 2 = 0 + 0.003 112 792 968 708 017 356 8;
  • 23) 0.003 112 792 968 708 017 356 8 × 2 = 0 + 0.006 225 585 937 416 034 713 6;
  • 24) 0.006 225 585 937 416 034 713 6 × 2 = 0 + 0.012 451 171 874 832 069 427 2;
  • 25) 0.012 451 171 874 832 069 427 2 × 2 = 0 + 0.024 902 343 749 664 138 854 4;
  • 26) 0.024 902 343 749 664 138 854 4 × 2 = 0 + 0.049 804 687 499 328 277 708 8;
  • 27) 0.049 804 687 499 328 277 708 8 × 2 = 0 + 0.099 609 374 998 656 555 417 6;
  • 28) 0.099 609 374 998 656 555 417 6 × 2 = 0 + 0.199 218 749 997 313 110 835 2;
  • 29) 0.199 218 749 997 313 110 835 2 × 2 = 0 + 0.398 437 499 994 626 221 670 4;
  • 30) 0.398 437 499 994 626 221 670 4 × 2 = 0 + 0.796 874 999 989 252 443 340 8;
  • 31) 0.796 874 999 989 252 443 340 8 × 2 = 1 + 0.593 749 999 978 504 886 681 6;
  • 32) 0.593 749 999 978 504 886 681 6 × 2 = 1 + 0.187 499 999 957 009 773 363 2;
  • 33) 0.187 499 999 957 009 773 363 2 × 2 = 0 + 0.374 999 999 914 019 546 726 4;
  • 34) 0.374 999 999 914 019 546 726 4 × 2 = 0 + 0.749 999 999 828 039 093 452 8;
  • 35) 0.749 999 999 828 039 093 452 8 × 2 = 1 + 0.499 999 999 656 078 186 905 6;
  • 36) 0.499 999 999 656 078 186 905 6 × 2 = 0 + 0.999 999 999 312 156 373 811 2;
  • 37) 0.999 999 999 312 156 373 811 2 × 2 = 1 + 0.999 999 998 624 312 747 622 4;
  • 38) 0.999 999 998 624 312 747 622 4 × 2 = 1 + 0.999 999 997 248 625 495 244 8;
  • 39) 0.999 999 997 248 625 495 244 8 × 2 = 1 + 0.999 999 994 497 250 990 489 6;
  • 40) 0.999 999 994 497 250 990 489 6 × 2 = 1 + 0.999 999 988 994 501 980 979 2;
  • 41) 0.999 999 988 994 501 980 979 2 × 2 = 1 + 0.999 999 977 989 003 961 958 4;
  • 42) 0.999 999 977 989 003 961 958 4 × 2 = 1 + 0.999 999 955 978 007 923 916 8;
  • 43) 0.999 999 955 978 007 923 916 8 × 2 = 1 + 0.999 999 911 956 015 847 833 6;
  • 44) 0.999 999 911 956 015 847 833 6 × 2 = 1 + 0.999 999 823 912 031 695 667 2;
  • 45) 0.999 999 823 912 031 695 667 2 × 2 = 1 + 0.999 999 647 824 063 391 334 4;
  • 46) 0.999 999 647 824 063 391 334 4 × 2 = 1 + 0.999 999 295 648 126 782 668 8;
  • 47) 0.999 999 295 648 126 782 668 8 × 2 = 1 + 0.999 998 591 296 253 565 337 6;
  • 48) 0.999 998 591 296 253 565 337 6 × 2 = 1 + 0.999 997 182 592 507 130 675 2;
  • 49) 0.999 997 182 592 507 130 675 2 × 2 = 1 + 0.999 994 365 185 014 261 350 4;
  • 50) 0.999 994 365 185 014 261 350 4 × 2 = 1 + 0.999 988 730 370 028 522 700 8;
  • 51) 0.999 988 730 370 028 522 700 8 × 2 = 1 + 0.999 977 460 740 057 045 401 6;
  • 52) 0.999 977 460 740 057 045 401 6 × 2 = 1 + 0.999 954 921 480 114 090 803 2;
  • 53) 0.999 954 921 480 114 090 803 2 × 2 = 1 + 0.999 909 842 960 228 181 606 4;
  • 54) 0.999 909 842 960 228 181 606 4 × 2 = 1 + 0.999 819 685 920 456 363 212 8;

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:


0.000 000 000 742 147 676 636 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 636 7(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:


0.000 000 000 742 147 676 636 7(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 -0.000 000 000 742 147 676 636 7 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