-0.000 000 000 742 147 676 661 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 661 4(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
-0.000 000 000 742 147 676 661 4(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:

|-0.000 000 000 742 147 676 661 4| = 0.000 000 000 742 147 676 661 4


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: 0.000 000 000 742 147 676 661 4.

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) 0.000 000 000 742 147 676 661 4 × 2 = 0 + 0.000 000 001 484 295 353 322 8;
  • 2) 0.000 000 001 484 295 353 322 8 × 2 = 0 + 0.000 000 002 968 590 706 645 6;
  • 3) 0.000 000 002 968 590 706 645 6 × 2 = 0 + 0.000 000 005 937 181 413 291 2;
  • 4) 0.000 000 005 937 181 413 291 2 × 2 = 0 + 0.000 000 011 874 362 826 582 4;
  • 5) 0.000 000 011 874 362 826 582 4 × 2 = 0 + 0.000 000 023 748 725 653 164 8;
  • 6) 0.000 000 023 748 725 653 164 8 × 2 = 0 + 0.000 000 047 497 451 306 329 6;
  • 7) 0.000 000 047 497 451 306 329 6 × 2 = 0 + 0.000 000 094 994 902 612 659 2;
  • 8) 0.000 000 094 994 902 612 659 2 × 2 = 0 + 0.000 000 189 989 805 225 318 4;
  • 9) 0.000 000 189 989 805 225 318 4 × 2 = 0 + 0.000 000 379 979 610 450 636 8;
  • 10) 0.000 000 379 979 610 450 636 8 × 2 = 0 + 0.000 000 759 959 220 901 273 6;
  • 11) 0.000 000 759 959 220 901 273 6 × 2 = 0 + 0.000 001 519 918 441 802 547 2;
  • 12) 0.000 001 519 918 441 802 547 2 × 2 = 0 + 0.000 003 039 836 883 605 094 4;
  • 13) 0.000 003 039 836 883 605 094 4 × 2 = 0 + 0.000 006 079 673 767 210 188 8;
  • 14) 0.000 006 079 673 767 210 188 8 × 2 = 0 + 0.000 012 159 347 534 420 377 6;
  • 15) 0.000 012 159 347 534 420 377 6 × 2 = 0 + 0.000 024 318 695 068 840 755 2;
  • 16) 0.000 024 318 695 068 840 755 2 × 2 = 0 + 0.000 048 637 390 137 681 510 4;
  • 17) 0.000 048 637 390 137 681 510 4 × 2 = 0 + 0.000 097 274 780 275 363 020 8;
  • 18) 0.000 097 274 780 275 363 020 8 × 2 = 0 + 0.000 194 549 560 550 726 041 6;
  • 19) 0.000 194 549 560 550 726 041 6 × 2 = 0 + 0.000 389 099 121 101 452 083 2;
  • 20) 0.000 389 099 121 101 452 083 2 × 2 = 0 + 0.000 778 198 242 202 904 166 4;
  • 21) 0.000 778 198 242 202 904 166 4 × 2 = 0 + 0.001 556 396 484 405 808 332 8;
  • 22) 0.001 556 396 484 405 808 332 8 × 2 = 0 + 0.003 112 792 968 811 616 665 6;
  • 23) 0.003 112 792 968 811 616 665 6 × 2 = 0 + 0.006 225 585 937 623 233 331 2;
  • 24) 0.006 225 585 937 623 233 331 2 × 2 = 0 + 0.012 451 171 875 246 466 662 4;
  • 25) 0.012 451 171 875 246 466 662 4 × 2 = 0 + 0.024 902 343 750 492 933 324 8;
  • 26) 0.024 902 343 750 492 933 324 8 × 2 = 0 + 0.049 804 687 500 985 866 649 6;
  • 27) 0.049 804 687 500 985 866 649 6 × 2 = 0 + 0.099 609 375 001 971 733 299 2;
  • 28) 0.099 609 375 001 971 733 299 2 × 2 = 0 + 0.199 218 750 003 943 466 598 4;
  • 29) 0.199 218 750 003 943 466 598 4 × 2 = 0 + 0.398 437 500 007 886 933 196 8;
  • 30) 0.398 437 500 007 886 933 196 8 × 2 = 0 + 0.796 875 000 015 773 866 393 6;
  • 31) 0.796 875 000 015 773 866 393 6 × 2 = 1 + 0.593 750 000 031 547 732 787 2;
  • 32) 0.593 750 000 031 547 732 787 2 × 2 = 1 + 0.187 500 000 063 095 465 574 4;
  • 33) 0.187 500 000 063 095 465 574 4 × 2 = 0 + 0.375 000 000 126 190 931 148 8;
  • 34) 0.375 000 000 126 190 931 148 8 × 2 = 0 + 0.750 000 000 252 381 862 297 6;
  • 35) 0.750 000 000 252 381 862 297 6 × 2 = 1 + 0.500 000 000 504 763 724 595 2;
  • 36) 0.500 000 000 504 763 724 595 2 × 2 = 1 + 0.000 000 001 009 527 449 190 4;
  • 37) 0.000 000 001 009 527 449 190 4 × 2 = 0 + 0.000 000 002 019 054 898 380 8;
  • 38) 0.000 000 002 019 054 898 380 8 × 2 = 0 + 0.000 000 004 038 109 796 761 6;
  • 39) 0.000 000 004 038 109 796 761 6 × 2 = 0 + 0.000 000 008 076 219 593 523 2;
  • 40) 0.000 000 008 076 219 593 523 2 × 2 = 0 + 0.000 000 016 152 439 187 046 4;
  • 41) 0.000 000 016 152 439 187 046 4 × 2 = 0 + 0.000 000 032 304 878 374 092 8;
  • 42) 0.000 000 032 304 878 374 092 8 × 2 = 0 + 0.000 000 064 609 756 748 185 6;
  • 43) 0.000 000 064 609 756 748 185 6 × 2 = 0 + 0.000 000 129 219 513 496 371 2;
  • 44) 0.000 000 129 219 513 496 371 2 × 2 = 0 + 0.000 000 258 439 026 992 742 4;
  • 45) 0.000 000 258 439 026 992 742 4 × 2 = 0 + 0.000 000 516 878 053 985 484 8;
  • 46) 0.000 000 516 878 053 985 484 8 × 2 = 0 + 0.000 001 033 756 107 970 969 6;
  • 47) 0.000 001 033 756 107 970 969 6 × 2 = 0 + 0.000 002 067 512 215 941 939 2;
  • 48) 0.000 002 067 512 215 941 939 2 × 2 = 0 + 0.000 004 135 024 431 883 878 4;
  • 49) 0.000 004 135 024 431 883 878 4 × 2 = 0 + 0.000 008 270 048 863 767 756 8;
  • 50) 0.000 008 270 048 863 767 756 8 × 2 = 0 + 0.000 016 540 097 727 535 513 6;
  • 51) 0.000 016 540 097 727 535 513 6 × 2 = 0 + 0.000 033 080 195 455 071 027 2;
  • 52) 0.000 033 080 195 455 071 027 2 × 2 = 0 + 0.000 066 160 390 910 142 054 4;
  • 53) 0.000 066 160 390 910 142 054 4 × 2 = 0 + 0.000 132 320 781 820 284 108 8;
  • 54) 0.000 132 320 781 820 284 108 8 × 2 = 0 + 0.000 264 641 563 640 568 217 6;

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:


0.000 000 000 742 147 676 661 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 661 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 676 661 4(10) =

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

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

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 661 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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