5.260 337 219 929 736 598 160 096 004 069 549 408 237 86 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.260 337 219 929 736 598 160 096 004 069 549 408 237 86(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
5.260 337 219 929 736 598 160 096 004 069 549 408 237 86(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. First, convert to binary (in base 2) the integer part: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.260 337 219 929 736 598 160 096 004 069 549 408 237 86.

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) 0.260 337 219 929 736 598 160 096 004 069 549 408 237 86 × 2 = 0 + 0.520 674 439 859 473 196 320 192 008 139 098 816 475 72;
  • 2) 0.520 674 439 859 473 196 320 192 008 139 098 816 475 72 × 2 = 1 + 0.041 348 879 718 946 392 640 384 016 278 197 632 951 44;
  • 3) 0.041 348 879 718 946 392 640 384 016 278 197 632 951 44 × 2 = 0 + 0.082 697 759 437 892 785 280 768 032 556 395 265 902 88;
  • 4) 0.082 697 759 437 892 785 280 768 032 556 395 265 902 88 × 2 = 0 + 0.165 395 518 875 785 570 561 536 065 112 790 531 805 76;
  • 5) 0.165 395 518 875 785 570 561 536 065 112 790 531 805 76 × 2 = 0 + 0.330 791 037 751 571 141 123 072 130 225 581 063 611 52;
  • 6) 0.330 791 037 751 571 141 123 072 130 225 581 063 611 52 × 2 = 0 + 0.661 582 075 503 142 282 246 144 260 451 162 127 223 04;
  • 7) 0.661 582 075 503 142 282 246 144 260 451 162 127 223 04 × 2 = 1 + 0.323 164 151 006 284 564 492 288 520 902 324 254 446 08;
  • 8) 0.323 164 151 006 284 564 492 288 520 902 324 254 446 08 × 2 = 0 + 0.646 328 302 012 569 128 984 577 041 804 648 508 892 16;
  • 9) 0.646 328 302 012 569 128 984 577 041 804 648 508 892 16 × 2 = 1 + 0.292 656 604 025 138 257 969 154 083 609 297 017 784 32;
  • 10) 0.292 656 604 025 138 257 969 154 083 609 297 017 784 32 × 2 = 0 + 0.585 313 208 050 276 515 938 308 167 218 594 035 568 64;
  • 11) 0.585 313 208 050 276 515 938 308 167 218 594 035 568 64 × 2 = 1 + 0.170 626 416 100 553 031 876 616 334 437 188 071 137 28;
  • 12) 0.170 626 416 100 553 031 876 616 334 437 188 071 137 28 × 2 = 0 + 0.341 252 832 201 106 063 753 232 668 874 376 142 274 56;
  • 13) 0.341 252 832 201 106 063 753 232 668 874 376 142 274 56 × 2 = 0 + 0.682 505 664 402 212 127 506 465 337 748 752 284 549 12;
  • 14) 0.682 505 664 402 212 127 506 465 337 748 752 284 549 12 × 2 = 1 + 0.365 011 328 804 424 255 012 930 675 497 504 569 098 24;
  • 15) 0.365 011 328 804 424 255 012 930 675 497 504 569 098 24 × 2 = 0 + 0.730 022 657 608 848 510 025 861 350 995 009 138 196 48;
  • 16) 0.730 022 657 608 848 510 025 861 350 995 009 138 196 48 × 2 = 1 + 0.460 045 315 217 697 020 051 722 701 990 018 276 392 96;
  • 17) 0.460 045 315 217 697 020 051 722 701 990 018 276 392 96 × 2 = 0 + 0.920 090 630 435 394 040 103 445 403 980 036 552 785 92;
  • 18) 0.920 090 630 435 394 040 103 445 403 980 036 552 785 92 × 2 = 1 + 0.840 181 260 870 788 080 206 890 807 960 073 105 571 84;
  • 19) 0.840 181 260 870 788 080 206 890 807 960 073 105 571 84 × 2 = 1 + 0.680 362 521 741 576 160 413 781 615 920 146 211 143 68;
  • 20) 0.680 362 521 741 576 160 413 781 615 920 146 211 143 68 × 2 = 1 + 0.360 725 043 483 152 320 827 563 231 840 292 422 287 36;
  • 21) 0.360 725 043 483 152 320 827 563 231 840 292 422 287 36 × 2 = 0 + 0.721 450 086 966 304 641 655 126 463 680 584 844 574 72;
  • 22) 0.721 450 086 966 304 641 655 126 463 680 584 844 574 72 × 2 = 1 + 0.442 900 173 932 609 283 310 252 927 361 169 689 149 44;
  • 23) 0.442 900 173 932 609 283 310 252 927 361 169 689 149 44 × 2 = 0 + 0.885 800 347 865 218 566 620 505 854 722 339 378 298 88;
  • 24) 0.885 800 347 865 218 566 620 505 854 722 339 378 298 88 × 2 = 1 + 0.771 600 695 730 437 133 241 011 709 444 678 756 597 76;

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).


4. 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:


0.260 337 219 929 736 598 160 096 004 069 549 408 237 86(10) =

0.0100 0010 1010 0101 0111 0101(2)

5. Positive number before normalization:

5.260 337 219 929 736 598 160 096 004 069 549 408 237 86(10) =

101.0100 0010 1010 0101 0111 0101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.260 337 219 929 736 598 160 096 004 069 549 408 237 86(10) =

101.0100 0010 1010 0101 0111 0101(2) =

101.0100 0010 1010 0101 0111 0101(2) × 20 =

1.0101 0000 1010 1001 0101 1101 01(2) × 22


7. 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 0 (a positive number)

Exponent (unadjusted): 2

Mantissa (not normalized):
1.0101 0000 1010 1001 0101 1101 01


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

2 + 2(8-1) - 1 =

(2 + 127)(10) =

129(10)


9. 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;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

129(10) =

1000 0001(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 010 1000 0101 0100 1010 1110 101 =

010 1000 0101 0100 1010 1110


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0001

Mantissa (23 bits) =
010 1000 0101 0100 1010 1110


Decimal number 5.260 337 219 929 736 598 160 096 004 069 549 408 237 86 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0001 - 010 1000 0101 0100 1010 1110

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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